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SAT Math : Algebra

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Example Questions

Example Question #36 : Quadratic Equations

Consider the equation

$4 x ^{2}+ 21 = 5x$ .

Which of the following statements correctly describes its solution set?

Possible Answers:

Exactly two solutions, both of which are rational.

Exactly one solution, which is irrational.

Exactly two solutions, both of which are irrational.

Exactly two solutions, both of which are imaginary.

Exactly one solution, which is rational.

Correct answer:

Exactly two solutions, both of which are imaginary.

Explanation:

Write the quadratic equation in standard form $ax^{2} + bx+ c = 0$ by subtracting from both sides:

$4 x ^{2}+ 21 = 5x$

$4 x ^{2}+ 21 - 5x = 5x - 5x$

$4 x ^{2}- 5x + 21 = 0$

The nature of the solution set of a quadratic equation in standard form can be determined by examining the discriminant $b^{2} - 4ac$ . Setting a = 4, b= -5 , c= 21 :

$b^{2} - 4ac =(-5) ^{2} - 4 \cdot 4 \cdot 21 = 25 - 336= -311$

The discriminant is negative, so there are two imaginary solutions.

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Example Question #27 : How To Find The Solution To A Quadratic Equation

Consider the equation

$2x ^{2} + 7x = 12$ .

Which of the following statements correctly describes its solution set?

Possible Answers:

Exactly one solution, which is rational.

Exactly two solutions, both of which are imaginary.

Exactly two solutions, both of which are irrational.

Exactly one solution, which is irrational.

Exactly two solutions, both of which are rational.

Correct answer:

Exactly two solutions, both of which are irrational.

Explanation:

Write the quadratic equation in standard form $ax^{2} + bx+ c = 0$ by subtracting 12 from both sides:

$2x ^{2} + 7x = 12$

$2x ^{2} + 7x - 12 = 12 - 12$

$2x ^{2} + 7x - 12 = 0$

The nature of the solution set of a quadratic equation in standard form can be determined by examining the discriminant $b^{2} - 4ac$ . Setting a = 2, b= 7, c= -12 :

$b^{2} - 4ac = 7 ^{2} - 4 \cdot 2 \cdot (-12) = 49 - (-96) = 145$

The discriminant is positive, so there are two real solutions. However, 145 is not a perfect square; the two solutions are therefore irrational.

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Example Question #341 : Equations / Inequalities

Consider the equation:

$3x^{2} + 7x =$ __________

Fill in the blank with a real constant to form an equation with exactly one real solution.

Possible Answers:

$- \frac{49}{12}$

None of the other responses gives a correct answer.

$- \frac{49}{36}$

Correct answer:

$- \frac{49}{12}$

Explanation:

We will call the constant that goes in the blank . The equation becomes

$3x^{2} + 7x = N$

Write the quadratic equation in standard form $ax^{2} + bx+ c = 0$ by subtracting from both sides:

$3x^{2} + 7x - N = N - N$

$3x^{2} + 7x - N = 0$

The solution set comprises exactly one rational solution if and only if the discriminant $b^{2} - 4ac$ is equal to 0. Setting a = 3, b= 7 , c= -N . and substituting in the equation:

$b^{2} - 4ac = 0$

$7^{2} - 4(3)(-N) = 0$

Solving for :

49 - (-12N) = 0

49 + 12 N = 0

49 + 12 N- 49 = 0 - 49

12N = -49

$12N\div 12 = -49 \div 12$

$N = - \frac{49}{12}$ ,

the correct response.

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Example Question #342 : Equations / Inequalities

Consider the equation:

__________ $x^{2} + 16 = 7x$

Fill in the blank with a real constant to form an equation with exactly one real solution.

Possible Answers:

$\frac{64}{49}$

$\frac{7}{8}$

None of the other responses gives a correct answer.

$\frac{8}{7}$

$\frac{49}{64}$

Correct answer:

$\frac{49}{64}$

Explanation:

We will call the constant that goes in the blank . The equation becomes

$N x^{2} + 16 = 7x$

Write the quadratic equation in standard form $ax^{2} + bx+ c = 0$ by subtracting from both sides:

$N x^{2} + 16 - 7x = 7x - 7x$

$N x^{2}- 7x + 16 = 0$

The solution set comprises exactly one rational solution if and only if the discriminant $b^{2} - 4ac$ is equal to 0. Setting a =N, b= -7 , c= 16 . and substituting in the equation:

$b^{2} - 4ac = 0$

$(-7)^{2} - 4 (N)(16)= 0$

49 - 64N= 0

Solving for :

49 - 64N+ 64N = 0 + 64N

49 = 64N

$49 \div 64 = 64N \div 64$

$N = \frac{49}{64}$ ,

the correct response.

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Example Question #341 : Algebra

Consider the equation:

$3 x^{2} + 48 =$ _________

Fill in the blank with a real constant to form an equation with exactly one real solution.

Possible Answers:

Correct answer:

Explanation:

We will call the constant that goes in the blank . The equation becomes

$3 x^{2} + 48 = Nx$

Write the quadratic equation in standard form $ax^{2} + bx+ c = 0$ by subtracting from both sides:

$3 x^{2} + 48 - Nx = Nx - Nx$

$3 x^{2} - Nx + 48 = 0$

The solution set comprises exactly one rational solution if and only if the discriminant $b^{2} - 4ac$ is equal to 0. Setting a = 3, b = -N , c= 48 . and substituting in the equation:

$b^{2} - 4ac = 0$

$(-N)^{2} - 4 (3)(48)= 0$

$N^{2} - 576= 0$

Solving for :

$N^{2} - 576+ 576 = 0+ 576$

$N^{2} = 576$

$N = \pm \sqrt{576}$

$N = \pm 24$ ,

that is, either N = 24 or N = -24 .

is not a choice, but 24 is; this is the correct response.

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Example Question #341 : Algebra

0.7 x + 0.4 y = 0.07

0.4 x+ 0.7y = -0.95

Evaluate .

Possible Answers:

x = 1.3

x = 1 3

x = 2.1

The system has no solution.

x = 21

Correct answer:

x = 1.3

Explanation:

Multiply both sides of the top equation by 7:

0.7 x + 0.4 y = 0.07

7(0.7 x + 0.4 y )= 7(0.07)

7(0.7 x )+7( 0.4 y )= 7(0.07)

4.9 x+2.8 y = 0.49

Multiply both sides of the bottom equation by :

0.4 x+ 0.7y = -0.95

-4 (0.4 x+ 0.7y )=-4 ( -0.95)

-4(0.4 x)+ (-4)(0.7y )=-4( -0.95)

-1.6x-2.8y = 3.8

Add both sides of the equations to eliminate the terms:

4.9 x+2.8 y = 0.49

$\underline{-1.6x-2.8y = 3.8}$

3.3x = 4.29

Solve for :

$3.3x \div 3.3 = 4.29 \div 3.3$

x = 1.3

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Example Question #41 : Quadratic Equations

Find the solutions of the equation 2x^2-6x-36 .

Possible Answers:

x=-3, x=-6

x=3,x=-6

x=6,x=-6

x=-3, x=6

x=3,x=6

Correct answer:

x=-3, x=6

Explanation:

This is a quadratic equation because the leading term is of degree ; hence, its solutions can be easily calculated via the quadratic formula:

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

In order to use the quadratic formula to solve a quadratic equation, you must identify the values of the coefficients , , and , substitute them into the quadratic formula, and perform the arithmetical calculations to yield one, two, or no real number solutions for .

In this case, a=2 , b=-6 , and c=-36 . Hence, the quadratic formula yields

$x=\frac{-(-6)\pm\sqrt{(-6)^2-4(2)(-36)}}{2(2)}$

$=\frac{6\pm\sqrt{36+288}}{4}$

$=\frac{6\pm\sqrt{324}}{4}$

$=\frac{6\pm18}{4}$

x=6, x=-3

Hence, this equation has two real solutions: and .

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Example Question #2 : Quadratic Equations

What is the sum of all the values of that satisfy:

3x^2-5x=2x+6

Possible Answers:

$\frac{7}{3}$

$\frac{2}{3}$

$\frac{14}{15}$

Correct answer:

$\frac{7}{3}$

Explanation:

With quadratic equations, always begin by getting it into standard form:

Ax^2+Bx+C=0

Therefore, take our equation:

3x^2-5x=2x+6

And rewrite it as:

3x^2-7x-6=0

You could use the quadratic formula to solve this problem. However, it is possible to factor this if you are careful. Factored, the equation can be rewritten as:

(3x+2)(x-3)=0

Now, either one of the groups on the left could be and the whole equation would be . Therefore, you set up each as a separate equation and solve for :

3x+2=0

3x=-2

$x=\frac{-2}{3}$

x-3=0

x=3

The sum of these values is:

$3+(-\frac{2}{3})=3-\frac{2}{3}=\frac{9}{3}-\frac{2}{3}=\frac{7}{3}$

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Example Question #41 : Quadratic Equations

Tommy throws a rock off a 10 meter ledge at a speed of 3 meters/second. Calculate when the rock hits the ground.

To solve use the equation

s(t)=-gt^2+v_0(t)+h_0

where

$\\g=4.9 \\v_0=\text{speed} \\h_0=\text{starting height}$

Possible Answers:

$1.5\text{ seconds}$

$1.25\text{ seconds}$

$1.75\text{ seconds}$

$2\text{ seconds}$

$1.3\text{ seconds}$

Correct answer:

$1.75\text{ seconds}$

Explanation:

Tommy throws a rock off a 10 meter ledge at a speed of 3 meters/second. To calculate when the rock hits the ground first identify what is known.

Using the equation

s(t)=-gt^2+v_0(t)+h_0

where

$\\g=4.9 \\v_0=\text{speed} \\h_0=\text{starting height}$

it is known that,

$v_0=3\frac{ \text{ m}}{\text{ s}}$

$h_0=10 \text{ m}$

Substituting the given values into the position s(t) equation looks as follows.

s(t)=-4.9t^2+3(t)+10

Now to calculate when the rock hits the ground, find the value that results in s(t)=0 .

0=-4.9t^2+3t+10

Use graphing technology to graph s(t) .

Screen shot 2016 02 11 at 8.18.52 am

It appears that the rock hits the ground approximately 1.75 seconds after Tommy throws it.

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Example Question #42 : Quadratic Equations

Find the solutions for

f(x)=x^2+qx+r

Possible Answers:

$x=\frac{-q\pm\sqrt{q^2-4 r}}{2}$

$x=\frac{-q\pm\sqrt{q^2-4 q}}{2}$

$x=\frac{q\pm\sqrt{q^2-4 r}}{2}$

$x=\frac{-r\pm\sqrt{r^2-4 q}}{2}$

Correct answer:

$x=\frac{-q\pm\sqrt{q^2-4 r}}{2}$

Explanation:

The first step is to set it equal to zero.

0=x^2+qx+r

Now we will use the quadratic formula.

$x=\frac{-b\pm\sqrt{b^2-4\cdot a\cdot c}}{2\cdot a}$

In this case a=1 , b=q , c=r

$x=\frac{-q\pm\sqrt{q^2-4\cdot 1\cdot r}}{2\cdot 1}$

$x=\frac{-q\pm\sqrt{q^2-4 r}}{2}$

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SAT Math : Algebra

Study concepts, example questions & explanations for SAT Math

All SAT Math Resources

Example Questions

Example Question #36 : Quadratic Equations

Example Question #27 : How To Find The Solution To A Quadratic Equation

Example Question #341 : Equations / Inequalities

Example Question #342 : Equations / Inequalities

Example Question #341 : Algebra

Example Question #341 : Algebra

Example Question #41 : Quadratic Equations

Example Question #2 : Quadratic Equations

Example Question #41 : Quadratic Equations

Example Question #42 : Quadratic Equations

All SAT Math Resources

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