Here at Zinkerz, we closely monitor trends in the SAT to bring our students the most up-to-date test-taking strategies. We try to get inside the minds of the test makers to help our students out-think the exam! Let’s review some algebra topics popping up on the SAT recently and tackle the best way to solve them!
Using the Discriminant
These questions will ask you to make a statement about the value of some unknown constant or coefficient in a given equation. To help you do this, the question will provide information about the number of solutions for that equation. This is a clue that will lead us to the best strategy for solving!
If a question tells you about the number of solutions for a given quadratic equation, you should immediately think of the discriminant. Recall:
![2 lines Line 1: discriminant equals b squared minus 4 a. c comma Line 2: for a. x squared plus b x plus c equals 0 <math xmlns="http://www.w3.org/1998/Math/MathML" display="block" data-is-equatio="1" data-latex="begin{array}{c}text{discriminant}=b^2-4ac,\
text{for} ax^2+bx+c=0end{array}"><mtable columnspacing="1em" rowspacing="4pt"><mtr><mtd><mtext>discriminant</mtext><mo>=</mo><msup><mi>b</mi><mn>2</mn></msup><mo>−</mo><mn>4</mn><mi>a</mi><mi>c</mi><mo>,</mo></mtd></mtr><mtr><mtd><mtext>for</mtext><mtext></mtext><mi>a</mi><msup><mi>x</mi><mn>2</mn></msup><mo>+</mo><mi>b</mi><mi>x</mi><mo>+</mo><mi>c</mi><mo>=</mo><mn>0</mn></mtd></mtr></mtable></math>](https://zinkerz.com/wp-content/uploads/2023/09/image-4.png)
The discriminant should look familiar! Recall the quadratic formula:
![x equals the fraction with numerator negative b plus or minus the square root of b squared minus 4 a. c and denominator 2 a. <math xmlns="http://www.w3.org/1998/Math/MathML" display="block" data-is-equatio="1" data-latex="x=frac{-bpmsqrt{b^2-4ac}}{2a}"><mi>x</mi><mo>=</mo><mfrac><mrow><mo>−</mo><mi>b</mi><mo>±</mo><msqrt><msup><mi>b</mi><mn>2</mn></msup><mo>−</mo><mn>4</mn><mi>a</mi><mi>c</mi></msqrt></mrow><mrow><mn>2</mn><mi>a</mi></mrow></mfrac></math>](https://zinkerz.com/wp-content/uploads/2023/09/image.png)
The discriminant comes directly from beneath the square root and tells us the following:
![3 lines Line 1: If b squared minus 4 a. c is less than 0 comma then no real number solutions exist period Line 2: If b squared minus 4 a. c equals 0 comma then there exists a. repeated real number solution period Line 3: If b squared minus 4 a. c is greater than 0 comma then two distinct real number solutions exist period <math xmlns="http://www.w3.org/1998/Math/MathML" display="block" data-is-equatio="1" data-latex="begin{array}{l}text{If} b^2-4ac<0, text{then no real number solutions exist.}\
text{If} b^2-4ac=0, text{then there exists a repeated real number solution.}\
text{If }b^2-4ac>0, text{then two distinct real number solutions exist.}end{array}"><mtable columnalign="left" columnspacing="1em" rowspacing="4pt"><mtr><mtd><mtext>If</mtext><mtext></mtext><msup><mi>b</mi><mn>2</mn></msup><mo>−</mo><mn>4</mn><mi>a</mi><mi>c</mi><mo><</mo><mn>0</mn><mo>,</mo><mtext></mtext><mtext>then no real number solutions exist.</mtext></mtd></mtr><mtr><mtd><mtext>If</mtext><mtext></mtext><msup><mi>b</mi><mn>2</mn></msup><mo>−</mo><mn>4</mn><mi>a</mi><mi>c</mi><mo>=</mo><mn>0</mn><mo>,</mo><mtext></mtext><mtext>then there exists a repeated real number solution.</mtext></mtd></mtr><mtr><mtd><mtext>If </mtext><msup><mi>b</mi><mn>2</mn></msup><mo>−</mo><mn>4</mn><mi>a</mi><mi>c</mi><mo>></mo><mn>0</mn><mo>,</mo><mtext></mtext><mtext>then two distinct real number solutions exist.</mtext></mtd></mtr></mtable></math>](https://zinkerz.com/wp-content/uploads/2023/09/image-3.png)
If you can remember these three conditions when calculating the discriminant, these problems will be easy to solve. Let’s do an example!
Problem 1:
![8 lines Line 1: x squared plus 12 x plus k equals 0 Line 2: In the equation above comma k comma is a. constant period If the equation has exactly two Line 3: distinct solutions comma which statement about the value of k must be true question mark Line 4: blank Line 5: cap A close paren k is greater than 36 Line 6: cap B close paren k equals 36 Line 7: cap C close paren k is less than 36 Line 8: cap D close paren k equals 12 <math xmlns="http://www.w3.org/1998/Math/MathML" display="block" data-is-equatio="1" data-latex="begin{array}{l} x^2+12x+k=0\
text{In the equation above,} k, text{is a constant. If the equation has exactly two }\
text{distinct solutions, which statement about the value of }k text{must be true?}\
\
text{ A) }k>36\
text{ B)} k=36\
text{ C) }k<36\
text{ D) }k=12end{array}"><mtable columnalign="left" columnspacing="1em" rowspacing="4pt"><mtr><mtd><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><msup><mi>x</mi><mn>2</mn></msup><mo>+</mo><mn>12</mn><mi>x</mi><mo>+</mo><mi>k</mi><mo>=</mo><mn>0</mn></mtd></mtr><mtr><mtd><mtext>In the equation above,</mtext><mtext></mtext><mi>k</mi><mo>,</mo><mtext></mtext><mtext>is a constant. If the equation has exactly two </mtext></mtd></mtr><mtr><mtd><mtext>distinct solutions, which statement about the value of </mtext><mi>k</mi><mtext></mtext><mtext>must be true?</mtext></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mtext> A) </mtext><mi>k</mi><mo>></mo><mn>36</mn></mtd></mtr><mtr><mtd><mtext> B)</mtext><mtext></mtext><mi>k</mi><mo>=</mo><mn>36</mn></mtd></mtr><mtr><mtd><mtext> C) </mtext><mi>k</mi><mo><</mo><mn>36</mn></mtd></mtr><mtr><mtd><mtext> D) </mtext><mi>k</mi><mo>=</mo><mn>12</mn></mtd></mtr></mtable></math>](https://zinkerz.com/wp-content/uploads/2023/09/image-5.png)
Step 1: Identify a, b, and c from the given equation.
![a. equals 1 comma b equals 12 comma c equals k <math xmlns="http://www.w3.org/1998/Math/MathML" display="block" data-is-equatio="1" data-latex="a=1, b=12, c=k"><mi>a</mi><mo>=</mo><mn>1</mn><mo>,</mo><mtext></mtext><mi>b</mi><mo>=</mo><mn>12</mn><mo>,</mo><mtext></mtext><mi>c</mi><mo>=</mo><mi>k</mi></math>](https://zinkerz.com/wp-content/uploads/2023/09/image-1.png)
Step 2: Because our equation has two distinct solutions, we will use the third condition of the discriminant to solve for k:
![5 lines Line 1: b squared minus 4 a. c is greater than 0 Line 2: 12 squared minus 4 times 1 times k is greater than 0 Line 3: 144 minus 4 k is greater than 0 Line 4: negative 4 k is greater than negative 144 Line 5: k is less than 36 <math xmlns="http://www.w3.org/1998/Math/MathML" display="block" data-is-equatio="1" data-latex="begin{array}{l} b^2-4ac>0\
12^2-4left(1right)left(kright)>0\
144-4k>0\
-4k>-144\
k<36end{array}"><mtable columnalign="left" columnspacing="1em" rowspacing="4pt"><mtr><mtd><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><msup><mi>b</mi><mn>2</mn></msup><mo>−</mo><mn>4</mn><mi>a</mi><mi>c</mi><mo>></mo><mn>0</mn></mtd></mtr><mtr><mtd><msup><mn>12</mn><mn>2</mn></msup><mo>−</mo><mn>4</mn><mrow data-mjx-texclass="INNER"><mo data-mjx-texclass="OPEN">(</mo><mn>1</mn><mo data-mjx-texclass="CLOSE">)</mo></mrow><mrow data-mjx-texclass="INNER"><mo data-mjx-texclass="OPEN">(</mo><mi>k</mi><mo data-mjx-texclass="CLOSE">)</mo></mrow><mo>></mo><mn>0</mn></mtd></mtr><mtr><mtd><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mn>144</mn><mo>−</mo><mn>4</mn><mi>k</mi><mo>></mo><mn>0</mn></mtd></mtr><mtr><mtd><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mo>−</mo><mn>4</mn><mi>k</mi><mo>></mo><mo>−</mo><mn>144</mn></mtd></mtr><mtr><mtd><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mi>k</mi><mo><</mo><mn>36</mn></mtd></mtr></mtable></math>](https://zinkerz.com/wp-content/uploads/2023/09/image-2.png)
Completing the Square
These questions will give you an equation for a circle and ask you to find the radius. This seems simple enough, except that the given equation is not in the form that is most conducive to recognizing the radius. These types of questions aim to get the given equation in standard circle form by completing the square. Recall the standard form for the equation of a circle:
![2 lines Line 1: open paren x minus h close paren squared plus open paren y minus k close paren squared equals r squared Line 2: where r is the radius and open paren h comma k close paren is the center of the circle <math xmlns="http://www.w3.org/1998/Math/MathML" display="block" data-is-equatio="1" data-latex="begin{array}{c}left(x-hright)^2+left(y-kright)^2=r^2\
text{where r is the radius and }left(h,kright) text{is the center of the circle}end{array}"><mtable columnspacing="1em" rowspacing="4pt"><mtr><mtd><msup><mrow data-mjx-texclass="INNER"><mo data-mjx-texclass="OPEN">(</mo><mi>x</mi><mo>−</mo><mi>h</mi><mo data-mjx-texclass="CLOSE">)</mo></mrow><mn>2</mn></msup><mo>+</mo><msup><mrow data-mjx-texclass="INNER"><mo data-mjx-texclass="OPEN">(</mo><mi>y</mi><mo>−</mo><mi>k</mi><mo data-mjx-texclass="CLOSE">)</mo></mrow><mn>2</mn></msup><mo>=</mo><msup><mi>r</mi><mn>2</mn></msup></mtd></mtr><mtr><mtd><mtext>where r is the radius and </mtext><mrow data-mjx-texclass="INNER"><mo data-mjx-texclass="OPEN">(</mo><mi>h</mi><mo>,</mo><mi>k</mi><mo data-mjx-texclass="CLOSE">)</mo></mrow><mtext></mtext><mtext>is the center of the circle</mtext></mtd></mtr></mtable></math>](https://zinkerz.com/wp-content/uploads/2023/09/image-6.png)
Once in this form, the circle’s radius is easily identifiable on the right side of the equation. Let’s look at an example, remembering that we will need to complete the square two times to achieve our goal.
Problem 2:
![9 lines Line 1: x squared plus y squared minus 12 x plus 16 y minus 21 equals 0 Line 2: blank Line 3: The equation of a. circle in the x y plane is shown above period Line 4: What is the radius of the circle question mark Line 5: blank Line 6: cap A close paren 2 Line 7: cap B close paren 8 Line 8: cap C close paren 11 Line 9: cap D close paren 12 <math xmlns="http://www.w3.org/1998/Math/MathML" display="block" data-is-equatio="1" data-latex="begin{array}{l} x^2+y^2-12x+16y-21=0\
\
text{The equation of a circle in the }xytext{ plane is shown above.}\
text{What is the radius of the circle?}\
\
text{A) }2\
text{B)} 8\
text{C) }11\
text{D)} 12\
end{array}"><mtable columnalign="left" columnspacing="1em" rowspacing="4pt"><mtr><mtd><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><msup><mi>x</mi><mn>2</mn></msup><mo>+</mo><msup><mi>y</mi><mn>2</mn></msup><mo>−</mo><mn>12</mn><mi>x</mi><mo>+</mo><mn>16</mn><mi>y</mi><mo>−</mo><mn>21</mn><mo>=</mo><mn>0</mn></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mtext>The equation of a circle in the </mtext><mi>x</mi><mi>y</mi><mtext> plane is shown above.</mtext></mtd></mtr><mtr><mtd><mtext>What is the radius of the circle?</mtext></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mtext>A) </mtext><mn>2</mn></mtd></mtr><mtr><mtd><mtext>B)</mtext><mtext></mtext><mn>8</mn></mtd></mtr><mtr><mtd><mtext>C) </mtext><mn>11</mn></mtd></mtr><mtr><mtd><mtext>D)</mtext><mtext></mtext><mn>12</mn></mtd></mtr></mtable></math>](https://zinkerz.com/wp-content/uploads/2023/09/image-7.png)
Let’s solve the following:
![Equatio Image <math xmlns="http://www.w3.org/1998/Math/MathML" data-is-equatio="1" data-equatio-url="https://equatio-api.texthelp.com/desmos/get/10-04-2023/2240d15a-f2f2-4857-bd61-69abca1260fd" data-equatio-type="image"><mglyph src="https://equatio-api.texthelp.com/desmos/get/10-04-2023/2240d15a-f2f2-4857-bd61-69abca1260fd" alt="Equatio Image"/></math>](https://zinkerz.com/wp-content/uploads/2023/05/image-8.png)
Picking the Correct Form
These questions will ask you to choose an equivalent form of the given equation where certain numbers appear in specific spots. The question will have hints about how to manipulate or solve the given equation to get it correct. Let’s look at an example:
Problem 3:
![10 lines Line 1: y equals 3 x squared minus 6 x plus 15 Line 2: blank Line 3: In which of the following equivalent forms of the given equation Line 4: do the coordinates of the vertex of its graph in the x y plane Line 5: appear as constants or coefficients question mark Line 6: blank Line 7: cap A close paren y equals 3 times open paren x squared minus 2 x plus 5 close paren Line 8: cap B close paren y equals 3 times open paren x squared minus 2 x close paren plus 12 Line 9: cap C close paren one third y minus 3 equals open paren x minus 5 close paren squared Line 10: cap D close paren y equals 3 times open paren x minus 1 close paren squared plus 12 <math xmlns="http://www.w3.org/1998/Math/MathML" display="block" data-is-equatio="1" data-latex="begin{array}{l} y=3x^2-6x+15\
\
text{In which of the following equivalent forms of the given equation}\
text{do the coordinates of the vertex of its graph in the }xy text{plane}\
text{appear as constants or coefficients?}\
\
text{A) }y=3left(x^2-2x+5right)\
text{B)} y=3left(x^2-2xright)+12\
text{C) }frac{1}{3}y-3=left(x-5right)^2\
text{D)} y=3left(x-1right)^2+12\
end{array}"><mtable columnalign="left" columnspacing="1em" rowspacing="4pt"><mtr><mtd><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mi>y</mi><mo>=</mo><mn>3</mn><msup><mi>x</mi><mn>2</mn></msup><mo>−</mo><mn>6</mn><mi>x</mi><mo>+</mo><mn>15</mn></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mtext>In which of the following equivalent forms of the given equation</mtext></mtd></mtr><mtr><mtd><mtext>do the coordinates of the vertex of its graph in the </mtext><mi>x</mi><mi>y</mi><mtext></mtext><mtext>plane</mtext></mtd></mtr><mtr><mtd><mtext>appear as constants or coefficients?</mtext></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mtext>A) </mtext><mi>y</mi><mo>=</mo><mn>3</mn><mrow data-mjx-texclass="INNER"><mo data-mjx-texclass="OPEN">(</mo><msup><mi>x</mi><mn>2</mn></msup><mo>−</mo><mn>2</mn><mi>x</mi><mo>+</mo><mn>5</mn><mo data-mjx-texclass="CLOSE">)</mo></mrow></mtd></mtr><mtr><mtd><mtext>B)</mtext><mtext></mtext><mi>y</mi><mo>=</mo><mn>3</mn><mrow data-mjx-texclass="INNER"><mo data-mjx-texclass="OPEN">(</mo><msup><mi>x</mi><mn>2</mn></msup><mo>−</mo><mn>2</mn><mi>x</mi><mo data-mjx-texclass="CLOSE">)</mo></mrow><mo>+</mo><mn>12</mn></mtd></mtr><mtr><mtd><mtext>C) </mtext><mfrac><mn>1</mn><mn>3</mn></mfrac><mi>y</mi><mo>−</mo><mn>3</mn><mo>=</mo><msup><mrow data-mjx-texclass="INNER"><mo data-mjx-texclass="OPEN">(</mo><mi>x</mi><mo>−</mo><mn>5</mn><mo data-mjx-texclass="CLOSE">)</mo></mrow><mn>2</mn></msup></mtd></mtr><mtr><mtd><mtext>D)</mtext><mtext></mtext><mi>y</mi><mo>=</mo><mn>3</mn><msup><mrow data-mjx-texclass="INNER"><mo data-mjx-texclass="OPEN">(</mo><mi>x</mi><mo>−</mo><mn>1</mn><mo data-mjx-texclass="CLOSE">)</mo></mrow><mn>2</mn></msup><mo>+</mo><mn>12</mn></mtd></mtr></mtable></math>](https://zinkerz.com/wp-content/uploads/2023/09/image-9.png)
The answer choices here may look a bit overwhelming, but a hint points us to the right one. The question asks us to look for the vertex coordinates in the equivalent form. We can use this vertex “clue” in a couple of ways:
Find the vertex (or part of it):
Remember that the equation for finding the x coordinate of the vertex point of a quadratic equation is:
![x equals negative b over 2 a. <math xmlns="http://www.w3.org/1998/Math/MathML" display="block" data-is-equatio="1" data-latex="x=-frac{b}{2a}"><mi>x</mi><mo>=</mo><mo>−</mo><mfrac><mi>b</mi><mrow><mn>2</mn><mi>a</mi></mrow></mfrac></math>](https://zinkerz.com/wp-content/uploads/2023/09/image-10.png)
Once you have the x coordinate, look for an answer choice where that number appears. Without also finding the y coordinate by plugging the x coordinate back into the original equation, we may only be able to narrow down our answer choices. By finding the x coordinate of the vertex, we will have one of the pieces we need to pick the right form.
Look for vertex form:
The question asks us to find the equivalent form with both coordinates of the vertex. Luckily, we know of a form that displays this: vertex form! Recall:
![2 lines Line 1: y equals a. times open paren x minus h close paren squared plus k Line 2: where open paren h comma k close paren is the vertex <math xmlns="http://www.w3.org/1998/Math/MathML" display="block" data-is-equatio="1" data-latex="begin{array}{c}y=aleft(x-hright)^2+k\
text{where }left(h,kright) text{is the vertex}end{array}"><mtable columnspacing="1em" rowspacing="4pt"><mtr><mtd><mi>y</mi><mo>=</mo><mi>a</mi><msup><mrow data-mjx-texclass="INNER"><mo data-mjx-texclass="OPEN">(</mo><mi>x</mi><mo>−</mo><mi>h</mi><mo data-mjx-texclass="CLOSE">)</mo></mrow><mn>2</mn></msup><mo>+</mo><mi>k</mi></mtd></mtr><mtr><mtd><mtext>where </mtext><mrow data-mjx-texclass="INNER"><mo data-mjx-texclass="OPEN">(</mo><mi>h</mi><mo>,</mo><mi>k</mi><mo data-mjx-texclass="CLOSE">)</mo></mrow><mtext></mtext><mtext>is the vertex</mtext></mtd></mtr></mtable></math>](https://zinkerz.com/wp-content/uploads/2023/09/image-11.png)
Look for an answer choice that is written in vertex form. This choice will have the coordinates of the vertex. You will know the value of a since it comes from the original equation, and h will be the x coordinate we found with the above formula.
You may only need to use one of these hints to be able to choose the right answer, so be sure to use your time efficiently. Between all these pieces of information, you can select the correct form!
Percent Change
These questions will ask you to write an expression representing a certain variable after some kind of percent change (increase or decrease). The trick to these questions is thinking of percents as whole numbers or decimals and understanding the algebraic relationship in the change of value. Let’s look at an example:
Problem 4:
![7 lines Line 1: The value of a. set of baseball cards comma b comma where b is greater than 0 comma has increased by 600 percent sign period Line 2: What is the resulting value in terms of b question mark Line 3: blank Line 4: cap A close paren 0.7 b Line 5: cap B close paren 7 b Line 6: cap C close paren 70 b Line 7: cap D close paren 700 b <math xmlns="http://www.w3.org/1998/Math/MathML" display="block" data-is-equatio="1" data-latex="begin{array}{l}text{The value of a set of baseball cards, }b, text{where} b>0, text{has increased by} 600%. \
text{What is the resulting value in terms of }b?\
\
text{A) }0.7b\
text{B) }7b\
text{C)} 70b\
text{D)} 700bend{array}"><mtable columnalign="left" columnspacing="1em" rowspacing="4pt"><mtr><mtd><mtext>The value of a set of baseball cards, </mtext><mi>b</mi><mo>,</mo><mtext></mtext><mtext>where</mtext><mtext></mtext><mi>b</mi><mo>></mo><mn>0</mn><mo>,</mo><mtext></mtext><mtext>has increased by</mtext><mtext></mtext><mn>600</mn><mi mathvariant="normal">%</mi><mo>.</mo><mtext></mtext></mtd></mtr><mtr><mtd><mtext>What is the resulting value in terms of </mtext><mi>b</mi><mo>?</mo></mtd></mtr><mtr><mtd/></mtr><mtr><mtd><mtext>A) </mtext><mn>0.7</mn><mi>b</mi></mtd></mtr><mtr><mtd><mtext>B) </mtext><mn>7</mn><mi>b</mi></mtd></mtr><mtr><mtd><mtext>C)</mtext><mtext></mtext><mn>70</mn><mi>b</mi></mtd></mtr><mtr><mtd><mtext>D)</mtext><mtext></mtext><mn>700</mn><mi>b</mi></mtd></mtr></mtable></math>](https://zinkerz.com/wp-content/uploads/2023/09/image-12-1024x313.png)
To calculate the “resulting value” of b, think about it as the sum of the “initial value” (or the original value, b) and the “additional value” (or the percent increase or decrease that has occurred). The initial value will be 100% of the value of b, or 1b, and the additional value will be 600% of the value of b, or 6b. Take a look at the equation below:
![3 lines Line 1: resulting value equals initial value plus additional value Line 2: resulting value equals 1 b plus 6 b Line 3: resulting value equals 7 b <math xmlns="http://www.w3.org/1998/Math/MathML" display="block" data-is-equatio="1" data-latex="begin{array}{l}text{resulting value}=text{initial value}+text{additional value}\
text{resulting value}=1b+6b\
text{resulting value}=7bend{array}"><mtable columnalign="left" columnspacing="1em" rowspacing="4pt"><mtr><mtd><mtext>resulting value</mtext><mo>=</mo><mtext>initial value</mtext><mo>+</mo><mtext>additional value</mtext></mtd></mtr><mtr><mtd><mtext>resulting value</mtext><mo>=</mo><mn>1</mn><mi>b</mi><mo>+</mo><mn>6</mn><mi>b</mi></mtd></mtr><mtr><mtd><mtext>resulting value</mtext><mo>=</mo><mn>7</mn><mi>b</mi></mtd></mtr></mtable></math>](https://zinkerz.com/wp-content/uploads/2023/09/image-13.png)
Don’t make these types of problems harder than they need to be! Oftentimes, the resulting equation is effortless to solve.
Rational Exponents and Radicals
These questions will ask you to choose the equivalent form of the given expression that contains rational exponents. This may require you to convert radicals to rational exponents or perform operations involving rational exponents to simplify. Sounds complicated? Remember these rules (tricks) to help you:
![2 lines Line 1: x to the a.-th power times x to the power of b equals x raised to the a. plus b power the fraction with numerator x to the a.-th power and denominator x to the power of b equals x raised to the a. minus b power Line 2: x to the a.-th power to the power of b equals x raised to the a. times b power x raised to the a. over b power equals the b-th root of x to the a.-th power <math xmlns="http://www.w3.org/1998/Math/MathML" display="block" data-is-equatio="1" data-latex="begin{array}{l}x^acdot x^b=x^{a+b}& frac{x^a}{x^b}=x^{a-b}\
left(x^aright)^b=x^{acdot b}& x^{frac{a}{b}}=sqrt[b]{x^a}end{array}"><mtable columnalign="left" columnspacing="1em" rowspacing="4pt"><mtr><mtd><msup><mi>x</mi><mi>a</mi></msup><mo>⋅</mo><msup><mi>x</mi><mi>b</mi></msup><mo>=</mo><msup><mi>x</mi><mrow data-mjx-texclass="ORD"><mi>a</mi><mo>+</mo><mi>b</mi></mrow></msup></mtd><mtd><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mfrac><msup><mi>x</mi><mi>a</mi></msup><msup><mi>x</mi><mi>b</mi></msup></mfrac><mo>=</mo><msup><mi>x</mi><mrow data-mjx-texclass="ORD"><mi>a</mi><mo>−</mo><mi>b</mi></mrow></msup></mtd></mtr><mtr><mtd><msup><mrow data-mjx-texclass="INNER"><mo data-mjx-texclass="OPEN">(</mo><msup><mi>x</mi><mi>a</mi></msup><mo data-mjx-texclass="CLOSE">)</mo></mrow><mi>b</mi></msup><mo>=</mo><msup><mi>x</mi><mrow data-mjx-texclass="ORD"><mi>a</mi><mo>⋅</mo><mi>b</mi></mrow></msup></mtd><mtd><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><msup><mi>x</mi><mrow data-mjx-texclass="ORD"><mfrac><mi>a</mi><mi>b</mi></mfrac></mrow></msup><mo>=</mo><mroot><msup><mi>x</mi><mi>a</mi></msup><mi>b</mi></mroot></mtd></mtr></mtable></math>](https://zinkerz.com/wp-content/uploads/2023/09/image-14.png)
These rules work when we simplify and perform operations on numbers with the same base. Furthermore, remember to follow the order of operations, so you don’t confuse yourself. Do one step at a time and keep your written work organized. Let’s try an example:
Problem 5:
![5 lines Line 1: Which expression is equivalent to b raised to the 11 over 12 power times b raised to the five fourths power raised to the exponent two thirds end exponent comma where b is greater than 0 question mark Line 2: cap A close paren the sixth root of b to the 11th power Line 3: cap B close paren the fourth root of b to the fifth power Line 4: cap C close paren the sixth root of b to the fifth power Line 5: cap D close paren the cube root of b to the fourth power <math xmlns="http://www.w3.org/1998/Math/MathML" display="block" data-is-equatio="1" data-latex="begin{array}{l}text{Which expression is equivalent to }b^{frac{11}{12}}left(b^{frac{5}{4}}right)^{frac{2}{3}},text{ where }b>0?\
text{A)} sqrt[6]{b^{11}}\
text{B)} sqrt[4]{b^5}\
text{C)} sqrt[6]{b^5}\
text{D)} sqrt[3]{b^4}end{array}"><mtable columnalign="left" columnspacing="1em" rowspacing="4pt"><mtr><mtd><mtext>Which expression is equivalent to </mtext><msup><mi>b</mi><mrow data-mjx-texclass="ORD"><mfrac><mn>11</mn><mn>12</mn></mfrac></mrow></msup><msup><mrow data-mjx-texclass="INNER"><mo data-mjx-texclass="OPEN">(</mo><msup><mi>b</mi><mrow data-mjx-texclass="ORD"><mfrac><mn>5</mn><mn>4</mn></mfrac></mrow></msup><mo data-mjx-texclass="CLOSE">)</mo></mrow><mrow data-mjx-texclass="ORD"><mfrac><mn>2</mn><mn>3</mn></mfrac></mrow></msup><mo>,</mo><mtext> where </mtext><mi>b</mi><mo>></mo><mn>0</mn><mo>?</mo></mtd></mtr><mtr><mtd><mtext>A)</mtext><mtext></mtext><mroot><msup><mi>b</mi><mrow data-mjx-texclass="ORD"><mn>11</mn></mrow></msup><mn>6</mn></mroot></mtd></mtr><mtr><mtd><mtext>B)</mtext><mtext></mtext><mroot><msup><mi>b</mi><mn>5</mn></msup><mn>4</mn></mroot></mtd></mtr><mtr><mtd><mtext>C)</mtext><mtext></mtext><mroot><msup><mi>b</mi><mn>5</mn></msup><mn>6</mn></mroot></mtd></mtr><mtr><mtd><mtext>D)</mtext><mtext></mtext><mroot><msup><mi>b</mi><mn>4</mn></msup><mn>3</mn></mroot></mtd></mtr></mtable></math>](https://zinkerz.com/wp-content/uploads/2023/09/image-15.png)
Notice how all answer choices are written as one term involving a radical? This is a cue for simplifying our given expression and converting our rational exponents. Since each term has the same base (b), we can follow the rules above to get the correct answer quickly.
![5 lines Line 1: b raised to the 11 over 12 power times b raised to the five fourths power raised to the exponent two thirds end exponent equals b raised to the 11 over 12 power times b raised to the 10 over 12 power Line 2: equals b raised to the 11 over 12 plus 10 over 12 power Line 3: equals b raised to the 22 over 12 power Line 4: equals b raised to the eleven sixths power Line 5: equals the sixth root of b to the 11th power <math xmlns="http://www.w3.org/1998/Math/MathML" display="block" data-is-equatio="1" data-latex="begin{array}{l}b^{frac{11}{12}}left(b^{frac{5}{4}}right)^{frac{2}{3}}=&b^{frac{11}{12}}cdot b^{frac{10}{12}}\
=&b^{frac{11}{12}+frac{10}{12}}\
=&b^{frac{22}{12}}\
=&b^{frac{11}{6}}\
=&sqrt[6]{b^{11}}end{array}"><mtable columnalign="left" columnspacing="1em" rowspacing="4pt"><mtr><mtd><msup><mi>b</mi><mrow data-mjx-texclass="ORD"><mfrac><mn>11</mn><mn>12</mn></mfrac></mrow></msup><msup><mrow data-mjx-texclass="INNER"><mo data-mjx-texclass="OPEN">(</mo><msup><mi>b</mi><mrow data-mjx-texclass="ORD"><mfrac><mn>5</mn><mn>4</mn></mfrac></mrow></msup><mo data-mjx-texclass="CLOSE">)</mo></mrow><mrow data-mjx-texclass="ORD"><mfrac><mn>2</mn><mn>3</mn></mfrac></mrow></msup><mo>=</mo></mtd><mtd><msup><mi>b</mi><mrow data-mjx-texclass="ORD"><mfrac><mn>11</mn><mn>12</mn></mfrac></mrow></msup><mo>⋅</mo><msup><mi>b</mi><mrow data-mjx-texclass="ORD"><mfrac><mn>10</mn><mn>12</mn></mfrac></mrow></msup></mtd></mtr><mtr><mtd><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mo>=</mo></mtd><mtd><msup><mi>b</mi><mrow data-mjx-texclass="ORD"><mfrac><mn>11</mn><mn>12</mn></mfrac><mo>+</mo><mfrac><mn>10</mn><mn>12</mn></mfrac></mrow></msup></mtd></mtr><mtr><mtd><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mo>=</mo></mtd><mtd><msup><mi>b</mi><mrow data-mjx-texclass="ORD"><mfrac><mn>22</mn><mn>12</mn></mfrac></mrow></msup></mtd></mtr><mtr><mtd><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mo>=</mo></mtd><mtd><msup><mi>b</mi><mrow data-mjx-texclass="ORD"><mfrac><mn>11</mn><mn>6</mn></mfrac></mrow></msup></mtd></mtr><mtr><mtd><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mtext></mtext><mo>=</mo></mtd><mtd><mroot><msup><mi>b</mi><mrow data-mjx-texclass="ORD"><mn>11</mn></mrow></msup><mn>6</mn></mroot></mtd></mtr></mtable></math>](https://zinkerz.com/wp-content/uploads/2023/09/image-16.png)
The Math section of the SAT doesn’t have to be scary! Practice the problems that give you the most trouble and master the foundational topics in algebra and geometry. Remember to use these tips and tricks to use the limited time efficiently.
You got this!