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High School Math : Quadratic Equations and Inequalities

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Example Questions

Example Question #51 : Intermediate Single Variable Algebra

Write an equation with the given roots:

$5 \pm 3i$

Possible Answers:

x^2-8x+34=0

x^2-8x+24=0

x^2-10x+34=0

x^2-10x+24=0

x^2-10x+14=0

Correct answer:

x^2-10x+34=0

Explanation:

To write an equation, find the sum and product of the roots. The sum is the negative coefficient of , and the product is the integer.

Sum: (5 + 3i)+(5-3i)=10

Product: $(5+3i)\cdot (5-3i)=25-9i^2=34$

Subtract the sum and add the product.

The equation is:

x^2-10x+34=0

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Example Question #51 : Intermediate Single Variable Algebra

Write an equation with the given roots:

$-\frac{3}{5}, \frac{1}{2}$

Possible Answers:

11x^2+x-3=0

10x^2+x-3=0

7x^2+x-3=0

8x^2+x-3=0

9x^2+x-3=0

Correct answer:

10x^2+x-3=0

Explanation:

To write an equation, find the sum and product of the roots. The sum is the negative coefficient of , and the product is the integer.

Sum: $(\frac{-3}{5})+(\frac{1}{2})=\frac{-1}{10}$

Product: $(\frac{-3}{5})\cdot (\frac{1}{2})=\frac{-3}{10}$

Subtract the sum and add the product.

The equation is:

$x^2 + \frac{1}{10}x-\frac{3}{10}=0$

Multiply the equation by :

10x^2+x-3=0

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Example Question #5 : Understanding Quadratic Roots

Write an equation with the given roots:

$4 \pm \sqrt{3}$

Possible Answers:

2x^2-9x+13=0

x^2-8x+12=0

x^2-9x+13=0

x^2-7x+13=0

x^2-8x+13=0

Correct answer:

x^2-8x+13=0

Explanation:

To write an equation, find the sum and product of the roots. The sum is the negative coefficient of , and the product is the integer.

Sum: $(4 + \sqrt{3})+(4 - \sqrt{3})=8$

Product: $(4 + \sqrt{3})\cdot (4 - \sqrt{3})=16-3=13$

Subtract the sum and add the product.

The equation is:

x^2 -8x+13=0

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Example Question #1 : Solving Quadratic Equations

Find the zeros.

$f(t)=t^{2}-4t-21$

Possible Answers:

x=7,-7

x=7,3

x=-7,3

x=3,-3

x=-3,7

Correct answer:

x=-3,7

Explanation:

Factor the equation to (t-7)(t+3) . Set both equal to zero and you get and . Remember, the zeros of an equation are wherever the function crosses the -axis.

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Example Question #2 : Solving Quadratic Equations

Find the zeros.

$f(x)=x^{2}-10x$

Possible Answers:

x=1,-1

x=0

x=10

x=0,10

x=-10

Correct answer:

x=0,10

Explanation:

Factor out an from the equation so that you have x(x-10) . Set and x-10 equal to . Your roots are and .

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Example Question #3 : Solving Quadratic Equations

Find the zeros.

$f(x)=\frac{1}{5}(x+9)^{2}(x^{2}-4)$

Possible Answers:

x=-2,2

x=-9

x=-3,3,-9

x=-9,-2,2

x=2,-9

Correct answer:

x=-9,-2,2

Explanation:

Set (x+9)^2 equal to zero and you get . Set (x^2-4) equal to zero as well and you get and because when you take a square root, your answer will be positive and negative.

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Example Question #1 : Solving Quadratic Equations

Find the zeros.

$f(x)=27x^{2}-45x-18$

Possible Answers:

$x=-\frac{1}{3},2$

$x=\frac{1}{3}$

$x=9,-\frac{1}{3},2$

x=-2,-9

$x=\frac{1}{3},2$

Correct answer:

$x=-\frac{1}{3},2$

Explanation:

Factor out a from the entire equation. After that, you get 9(3x^2-5x-2) . Factor the expression to (3x+1)(x-2) . Set both of those equal to zero and your answers are $-\frac{1}{3}$ and .

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Example Question #4 : Solving Quadratic Equations

Find the zeros.

$f(x)=25x^{2}-81$

Possible Answers:

$x=-\frac{9}{5}$

x=9,-9

$x=\frac{9}{5},-\frac{9}{5}$

x=5

$x=\frac{9}{5}$

Correct answer:

$x=\frac{9}{5},-\frac{9}{5}$

Explanation:

This expression is the difference of perfect squares. Therefore, it factors to (5x-9)(5x+9) . Set both of those equal to zero and your answers are $\frac{9}{5}$ and $-\frac{9}{5}$ .

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Example Question #1 : Solving Quadratic Equations

Find the zeros.

f(x)= 36x^2-9x

Possible Answers:

x=8

x=0,4

$x=0,\frac{1}{4}$

x=0

$x=\frac{1}{4}$

Correct answer:

$x=0,\frac{1}{4}$

Explanation:

Factor the equation to 9x(4x-1) . Set both equal to and you get and $\frac{1}{4}$ .

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Example Question #6 : Solving Quadratic Equations

Find the zeros.

$f(x)=\frac{1}{4}x^2 +\frac{5}{4}x+\frac{3}{2}$

Possible Answers:

x=4,2

x=3,2

x=-3,3

x=2

x=-3,-2

Correct answer:

x=-3,-2

Explanation:

Factor a $\frac{1}{4}$ out of the quation to get

$\frac{1}{4}\left ( x^{2}+5x+6 \right )$

which can be further factored to

$\frac{1}{4}(x+3)(x+2)$ .

Set the last two expressions equal to zero and you get and .

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All High School Math Resources

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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 #51 : Intermediate Single Variable Algebra

Example Question #51 : Intermediate Single Variable Algebra

Example Question #5 : Understanding Quadratic Roots

Example Question #1 : Solving Quadratic Equations

Example Question #2 : Solving Quadratic Equations

Example Question #3 : Solving Quadratic Equations

Example Question #1 : Solving Quadratic Equations

Example Question #4 : Solving Quadratic Equations

Example Question #1 : Solving Quadratic Equations

Example Question #6 : Solving Quadratic Equations

All High School Math Resources

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