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High School Math : Quadratic Equations and Inequalities

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Example Questions

Example Question #1 : Solving Quadratic Equations

Find the zeros.

f(x)=x(x-6)^2

Possible Answers:

x=0

x=0,6

x=1,-1

x=6

x=-6,6

Correct answer:

x=0,6

Explanation:

Set each expression equal to zero and you get 0 and 6.

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Example Question #2 : Solving Quadratic Equations

Find the zeros.

f(x)=(x-2)(x+4)^3

Possible Answers:

x=2

x=0,2,-2

x=4

x=2,-4

x=-2,4

Correct answer:

x=2,-4

Explanation:

Set both expressions equal to . The first factor yields . The second factor gives you .

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Example Question #8 : Finding Roots

Find the zeros.

f(x)=(x+5)(x-8)^2

Possible Answers:

x=-5,8

x=5,8

x=-8

x=0

x=-5

Correct answer:

x=-5,8

Explanation:

Set both expressions to and you get and .

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Example Question #61 : Intermediate Single Variable Algebra

Solve the following equation by factoring.

x^2+7x-8=0

Possible Answers:

$x=8\ or\ -1$

$x=-8\ or\ -1$

$x=-8\ or\ 1$

$x=8\ or\ 1$

Correct answer:

$x=-8\ or\ 1$

Explanation:

We can factor by determining the terms that will multiply to –8 and add to +7.

$8*-1=-8\ and\ 8+(-1)=7$

Our factors are +8 and –1.

0=x^2+7x-8=(x+8)(x-1)

Now we can set each factor equal to zero and solve for the root.

$x+8=0\ or\ x-1=0$

$x=-8\ or\ x=1$

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Example Question #62 : Intermediate Single Variable Algebra

Solve the following equation by factoring.

2x^2-8x-10=0

Possible Answers:

$x=5\ or\ 1$

$x=-5\ or\ -1$

$x=5\ or\ -1$

$x=-5\ or\ 1$

Correct answer:

$x=5\ or\ -1$

Explanation:

We know that one term has a coefficient of 2 and that our factors must multiply to –10.

$2*-5=-10\ and\ 2+2(-5)=-8$

Our factors are +2 and –5.

0=2x^2-8x-10=(2x+2)(x-5)

Now we can set each factor equal to zero and solve for the root.

$2x+2=0\ or\ x-5=0$

$x=-1\ or\ 5$

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Example Question #63 : Intermediate Single Variable Algebra

Solve the following equation by factoring.

0=x^3+4x^2+4x

Possible Answers:

$x=0\ or\ 2$

$x= 0,2\ or\ -2$

$x=0\ or\ -2$

$x=0,-4\ or\ 1$

Correct answer:

$x=0\ or\ -2$

Explanation:

First, we can factor an term out of all of the values.

0=x^3+4x^2+4x=x(x^2+4x+4)

We can factor remaining polynomial by determining the terms that will multiply to +4 and add to +4.

$2*2=4\ and\ 2+2=4$

Our factors are +2 and +2.

0=x(x+2)(x+2)=x(x+2)^2

Now we can set each factor equal to zero and solve for the root.

$x=0\ or\ (x+2)^2=0$

$x=0\ or\ -2$

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Example Question #71 : Intermediate Single Variable Algebra

Solve $x^{2}+6x+5=0$

Possible Answers:

$x=-5\; or\; x=-1$

$x=-5\; or\; x=-2$

$x=3\; or\; x=-1$

$x=2\; or\; x=3$

$x=-3\; or\; x=-2$

Correct answer:

$x=-5\; or\; x=-1$

Explanation:

Factor the problem and set each factor equal to zero.

$x^{2}+6x+5=0$ becomes (x+5)(x+1)=0 so $x=-5 \; or\; x=-1$

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Example Question #72 : Intermediate Single Variable Algebra

Solve $x^{2}-8x+12=0$ .

Possible Answers:

$x=6\; or\; x=-8$

$x=-2\; or\; x=2$

$x=6\; or\; x=2$

$x=-4\; or\; x=-2$

$x=6\; or\; x=-8$

$x=6\; or\; x=-4$

$x=-2\; or\; x=2$

Correct answer:

$x=6\; or\; x=2$

Explanation:

Factor the quadratic equation and set each factor equal to zero:

$x^{2}-8x+12=0$ becomes (x-6)(x-2)=0 so the correct answer is $x=6\; or \; x=2$ .

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Example Question #32 : Quadratic Equations And Inequalities

What are the roots of (x+3)(x-4) ?

Possible Answers:

(3,4)(4,3)

(-3,0)(4,0)

(0,3)(0,4)

(0,-3)(0,-4)

(3,0)(-4,0)

Correct answer:

(-3,0)(4,0)

Explanation:

To find the roots, we need to find the values that would make x=0 . Since there are two parts to (x+3)(x-4) , we will have two roots: one where x+3=0 , and one where x-4=0 .

Solve each one individually:

x+3=0

x=-3

x-4=0

x=4

Therefore, our roots will be (-3,0)(4,0) .

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Example Question #74 : Intermediate Single Variable Algebra

What are the roots of (x+5)(x-1) ?

Possible Answers:

(5,0)(-1,0)

(0,5)(0,-1)

$(\frac{1}{5},0)$

(0,-5)(0,1)

(-5,0)(1,0)

Correct answer:

(-5,0)(1,0)

Explanation:

To find the roots, we need to find what would make x=0 . Since there are two parts to (x+5)(x-1) , we will have two roots: one where x+5=0 , and one where x-1=0 .

Solve each individually.

x+5=0

x=-5

x-1=0

x=1

Our two roots will be (-5,0)(1,0) .

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High School Math : Quadratic Equations and Inequalities

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

Example Question #1 : Solving Quadratic Equations

Example Question #2 : Solving Quadratic Equations

Example Question #8 : Finding Roots

Example Question #61 : Intermediate Single Variable Algebra

Example Question #62 : Intermediate Single Variable Algebra

Example Question #63 : Intermediate Single Variable Algebra

Example Question #71 : Intermediate Single Variable Algebra

Example Question #72 : Intermediate Single Variable Algebra

Example Question #32 : Quadratic Equations And Inequalities

Example Question #74 : Intermediate Single Variable Algebra

All High School Math Resources

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