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SAT Math : How to find the solution to a quadratic equation

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Example Questions

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Example Question #11 : Quadratic Equation

The expression $x^{2} - 8x +12$ is equal to 0 when $x = 2$ and $x = ?$

Possible Answers:

$-6$

$4$

$-2$

$-12$

$6$

Correct answer:

$6$

Explanation:

Factor the expression and set each factor equal to 0:

$(x-2)(x-6)= 0$

$x-2 = 0$

$x = 2$

$x-6 = 0$

$x = 6$

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Example Question #11 : How To Find The Solution To A Quadratic Equation

Two positive consecutive multiples of four have a product of 96. What is the sum of the two numbers?

Possible Answers:

$16$

$20$

$12$

$8$

$24$

Correct answer:

$20$

Explanation:

Let $x$ = the first number and $x+4$ = the second number.

So the equation to solve becomes $x(x+4)=96$ . This quadratic equation needs to be multiplied out and set equal to zero before factoring. Then set each factor equal to zero and solve. Only positive numbers are correct, so the answer is $8+12=20$ .

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Example Question #23 : Quadratic Equations

Two consecutive positive odd numbers have a product of 35. What is the sum of the two numbers?

Possible Answers:

Correct answer:

Explanation:

Let = first positive number and x+2 = second positive number.

The equation to solve becomes

x(x+2)=35

We multiply out this quadratic equation and set it equal to 0, then factor.

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Example Question #21 : How To Find The Solution To A Quadratic Equation A1

The product of two consecutive positive multiples of four is 192. What is the sum of the two numbers?

Possible Answers:

Correct answer:

Explanation:

Let = the first positive number and x+4 = the second positive number

The equation to solve becomes x(x+4)=192

We solve this quadratic equation by multiplying it out and setting it equal to 0. The next step is to factor.

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Example Question #31 : How To Find The Solution To A Quadratic Equation A1

Two consecutive positive multiples of three have a product of 504. What is the sum of the two numbers?

Possible Answers:

Correct answer:

Explanation:

Let = the first positive number and x + 3 = the second positive number.

So the equation to solve is x(x + 3) = 504

We multiply out the equation and set it equal to zero before factoring.

$x^{2} + 3x - 504 = 0$ thus the two numbers are 21 and 24 for a sum of 45.

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Example Question #21 : Quadratic Equations

Two consecutive positive numbers have a product of 420. What is the sum of the two numbers?

Possible Answers:

Correct answer:

Explanation:

Let = first positive number and x + 1 = second positive number

So the equation to solve becomes x(x + 1) = 420

Using the distributive property, multiply out the equation and then set it equal to 0. Next factor to solve the quadratic.

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Example Question #21 : Quadratic Equations

A rectangle has a perimeter of $50\ m$ and an area of $150\ m^{2}$ What is the difference between the length and width?

Possible Answers:

$5\ m$

$10\ m$

$25\ m$

$20\ m$

$15\ m$

Correct answer:

$5\ m$

Explanation:

For a rectangle, $P=2w+2l$ and $A=lw$ where $w$ = width and $l$ = length.

So we get two equations with two unknowns:

$50=2w+2l$

$25=w+l$

$l=25-w$

$150=lw$

Making a substitution we get

$150=(25-w)w$

$w^{2} -25w + 150 = 0$

Solving the quadratic equation we get $w=10\ m$ or $15\ m$ .

$l=15\ m\ or\ 10\ m$

The difference is $5\ m$ .

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Example Question #28 : Quadratic Equations

If $x(x^2-5x+3)=-9, \ x>0,\ and \ y^2+y=2$ then which of the following is a possible value for xy^2+y ?

Possible Answers:

Correct answer:

Explanation:

$x(x^2-5x+3)=-9, \ x>0,\ and \ y^2+y=2$

$x(x^2-5x+3)=-9 \Rightarrow x^3-5x^2+3x + 9=0\Rightarrow (x^2-6x+9)(x+1)$

$(x^2-6x+9)(x+1) = (x-3)^2(x+1)\Rightarrow x= -1\ or\ 3$

Since x>0 , x=3 .

$y^2+y-2=0 \Rightarrow (y+2)(y-1)=0 \Rightarrow y=-2\ or\ 1.$

Thus $xy^2+y = (3)(-2)^2+(-2)\ or\ (3)(1)^2+(1) = 12-2\ or\ 3+1\ = 10\ or\ 4$

Of these two, only 4 is a possible answer.

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Example Question #29 : Quadratic Equations

Find all real solutions to the equation.

$x{^{2}+6x-16=0$

Possible Answers:

$4\ \text{or}\ -4$

$2\ \text{or}\ -8$

$-2\ \text{or}\ -8$

Correct answer:

$2\ \text{or}\ -8$

Explanation:

$x{^{2}+6x-16=0$

To solve by factoring, we need two numbers that add to $\small 6$ and multiply to $\small -16$ .

$-2*8=-16\ \text{and}\ -2+8=6$

(x-2)(x+8)=0

In order for the equation to equal zero, one of the terms must be equal to zero.

x-2=0

x=2

x+8=0

x=-8

Our final answer is that $x\epsilon\left \{ 2,-8 \right \}$ .

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Example Question #30 : Quadratic Equations

How many real solutions are there for the following equation?

x^4 + 5x^2 - 14 = 0

Possible Answers:

Correct answer:

Explanation:

The first thing to notice is that you have powers with a regular sequence. This means you can simply treat it like a quadratic equation. You are then able to factor it as follows:

The factoring can quickly be done by noticing that the 14 must be either 7*2 or 14*1 . Because it is negative, one constant will be negative and the other positive. Finally, since the difference between 14 and 1 cannot be 5, it must be 7 and 2.

Alternatively, one could use the quadratic formula.

The end result is that you have:

$x^2 + 7 = 0\Rightarrow x^2 = -7 \ or \ x^2-2 = 0\Rightarrow x^2=2$

$x^2=2 \Rightarrow x=\pm \sqrt{2}$

$x^2=-7\Rightarrow x=\pm \sqrt-7$

The latter of these two gives only complex answers, so there are two real answers.

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SAT Math : How to find the solution to a quadratic equation

Study concepts, example questions & explanations for SAT Math

All SAT Math Resources

Example Questions

Example Question #11 : Quadratic Equation

Example Question #11 : How To Find The Solution To A Quadratic Equation

Example Question #23 : Quadratic Equations

Example Question #21 : How To Find The Solution To A Quadratic Equation A1

Example Question #31 : How To Find The Solution To A Quadratic Equation A1

Example Question #21 : Quadratic Equations

Example Question #21 : Quadratic Equations

Example Question #28 : Quadratic Equations

Example Question #29 : Quadratic Equations

Example Question #30 : Quadratic Equations

All SAT Math Resources

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