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High School Math : Solving Quadratic Equations

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Example Questions

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 #11 : Finding Roots

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

Possible Answers:

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

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

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

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

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

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 #34 : Quadratic Equations And Inequalities

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

Possible Answers:

(3,0)(-4,0)

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

(3,4)(4,3)

(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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Example Question #72 : Intermediate Single Variable Algebra

What are the roots of (x+7)(x-5) ?

Possible Answers:

(-7,0)(5,0)

(7,0)(-5,0)

(0,7)(0,-5)

(0,-7)(0,5)

$(\frac{5}{7},0)(\frac{7}{5},0)$

Correct answer:

(-7,0)(5,0)

Explanation:

To find the roots, we need to find what would make x=0 . Since there are two parts to (x+7)(x-5) , we will have two roots: one where x+7=0 , and one where x-5=0 .

Solve each individually.

x+7=0

x=-7

x-5=0

x=5

Therefore, our two roots will be at (-7,0)(5,0) .

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Example Question #72 : Intermediate Single Variable Algebra

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

Possible Answers:

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

$x=-3\; \; or\; \; x=4$

$x=3\; \; or\; \; x=4$

No solutions

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

Correct answer:

$x=3\; \; or\; \; x=4$

Explanation:

Factor the equation by looking for two factors that multiply to +12 and add to .

The factors are and , so the equation to solve becomes (x-3)(x-4)=0 .

Next, set each factor equal to zero and solve:

x-3=0 or x-4=0

The solution is x=3 or x=4 .

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Example Question #41 : Quadratic Equations And Inequalities

Solve 2x^2-3x+1 .

Possible Answers:

x=-1, 1

x=10, 1

$x=1, \frac{1}{2}$

$x=\frac{1}{2}, -\frac{1}{2}$

x=1

Correct answer:

$x=1, \frac{1}{2}$

Explanation:

To find the roots of this equation, you can factor it to

(2x-1)(x-1)

Set each of those expressions equal to zero and then solve for . The roots are and $\frac{1}{2}$ .

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High School Math : Solving Quadratic Equations

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

Example Question #61 : Intermediate Single Variable Algebra

Example Question #62 : Intermediate Single Variable Algebra

Example Question #63 : Intermediate Single Variable Algebra

Example Question #11 : Finding Roots

Example Question #72 : Intermediate Single Variable Algebra

Example Question #34 : Quadratic Equations And Inequalities

Example Question #74 : Intermediate Single Variable Algebra

Example Question #72 : Intermediate Single Variable Algebra

Example Question #72 : Intermediate Single Variable Algebra

Example Question #41 : Quadratic Equations And Inequalities

All High School Math Resources

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