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High School Math : Solving Quadratic Equations

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Example Questions

Example Question #44 : Quadratic Equations And Inequalities

Find the root(s) of the following quadratic polynomial.

$f(x) = x^{2} + 4x + 4$

Possible Answers:

x=0

x=2

x=2, x=-2

x =- 2

Correct answer:

x =- 2

Explanation:

$f(x) = x^{2} + 4x + 4$

We set the function equal to 0 and factor the equation. By FOIL, we can confirm that (x+2)(x+2) = 0 is equivalent to the given function. Thus, the only zero comes from x + 2 = 0 , and x = -2 . Thus, x = -2 is the only root.

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Example Question #45 : Quadratic Equations And Inequalities

$\textup{What are the possible values of }x \textup{ when }\; \; x^{2}-2x-15=0\:\textup{?}$

Possible Answers:

3,5

-2, -15

-3, 5

-5. 3

3, 5 -5, -3

Correct answer:

-3, 5

Explanation:

$\textup{Factor to solve. The product of the constants in the two roots is -15, while }$

$\textup{their sum is -2. }\:\:\left ( x+3\right )\left ( x-5\right )=0\;\;\;\;x=-3 \textup{ or }x=5$

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Example Question #46 : Quadratic Equations And Inequalities

Solve the quadratic equation using any method:

bx^2+cx+a=0

Possible Answers:

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

$x=\frac{-c \pm \sqrt{c^2-2ab}}{2b}$

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

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

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

Correct answer:

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

Explanation:

Use the quadratic formula to solve:

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

a = b

b = c

c = a

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

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

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Example Question #81 : Intermediate Single Variable Algebra

Solve the following equation using the quadratic form:

$x-4\sqrt{x}-32=0$

Possible Answers:

x=49

x=64

x=36

x=16

x=25

Correct answer:

x=64

Explanation:

Factor and solve:

$x-4\sqrt{x}-32=0$

$(\sqrt{x}-8)(\sqrt{x}+4)=0$

$\sqrt{x}=8$

x=64

$\sqrt{x}=-4$

This has no solutions.

Therefore there is only one solution:

x=64

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Example Question #82 : Intermediate Single Variable Algebra

Solve the following equation using the quadratic form:

x^4-12x^2+27=0

Possible Answers:

$x=\pm 5, \pm \sqrt{5}$

$x=\pm 6, \pm \sqrt{6}$

$x=\pm 2, \pm \sqrt{2}$

$x=\pm 3, \pm \sqrt{3}$

$x=\pm 7, \pm \sqrt{7}$

Correct answer:

$x=\pm 3, \pm \sqrt{3}$

Explanation:

Factor and solve:

x^4-12x^2+27=0

(x^2-9)(x^2-3)=0

x^2-9=0

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

$x= \pm3$

x^2-3=0

x^2=3

$x=\pm \sqrt{3}$

Therefore the equation has four solutions:

$x= \pm3, \pm \sqrt{3}$

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Example Question #83 : Intermediate Single Variable Algebra

Solve the following equation using the quadratic form:

$x^{\frac{2}{3}}-9x^{\frac{1}{3}}+20=0$

Possible Answers:

x=124, 65

x=125, 64

x=120, 64

x=125, 65

x=124, 64

Correct answer:

x=125, 64

Explanation:

Factor and solve:

$x^{\frac{2}{3}}-9x^{\frac{1}{3}}+20=0$

$(x^{\frac{1}{3}})^2-9x^{\frac{1}{3}}+20=0$

$(x^\frac{1}{3}-5)(x^\frac{1}{3}-4)=0$

$x^\frac{1}{3}=5$

x=125

$x^\frac{1}{3}=4$

x=64

Therefore the equation has two solutions.

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Example Question #84 : Intermediate Single Variable Algebra

Solve the following equation using the quadratic form:

n^6+9n^3+8=0

Possible Answers:

$n=-3,-1,1 \pm i\sqrt{3}, \frac{1 \pm i\sqrt{3}}{2}$

$n=-3,-2,1 \pm i\sqrt{3}, \frac{1 \pm i\sqrt{3}}{2}$

$n=-3,-2,2 \pm i\sqrt{3}, \frac{1 \pm i\sqrt{3}}{2}$

$n=-2,-1,1 \pm i\sqrt{3}, \frac{1 \pm i\sqrt{3}}{2}$

$n=-3,-1,1 \pm i\sqrt{3}, \frac{3 \pm i\sqrt{3}}{2}$

Correct answer:

$n=-2,-1,1 \pm i\sqrt{3}, \frac{1 \pm i\sqrt{3}}{2}$

Explanation:

Factor and solve:

n^6+9n^3+8=0

(n^3)^2+9n^3+8=0

(n^3+8)(n^3+1)=0

(n+2)(n^2-2n+4)(n+1)(n^2-n+1)=0

Each of these factors gives solutions to the equation:

n=-2

n=-1

$n=1 \pm i\sqrt{3}$

$n=\frac{1 \pm i\sqrt{3}}{2}$

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Example Question #23 : Finding Roots

The product of two consecutive positive numbers is 156 . What is the sum of the two numbers?

Possible Answers:

Correct answer:

Explanation:

Let the first number and x+1= the second number.

The equation to sovle becomes x(x+1)=156 , or $x^{2}+x-156=0$ .

Factoring we get (x+13)(x-12)=0 , so the solution is $x=-13 \; \; or\; \; x=12$ . The problem states that the numbers are positive, so the correct numbers are and , which sum to .

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Example Question #332 : Algebra Ii

Two positive, consecutive odd numbers have a product of 143 . What is their sum?

Possible Answers:

Correct answer:

Explanation:

Let first odd number and x+2= second odd number. Then:

x(x+2)=143

Use the distributive property and subtract 143 from both sides to get $x^{2}+2x-143=0$ .

Factoring we get (x+13)(x-11)=0 .

Solving we get $x+13=0\; or\; x-11=0$ , so $x=-13\; or\; x=11$ .

The problem stated that the numbers were positive so the answer becomes 11+13=24 .

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Example Question #333 : Algebra Ii

Find the sum of the solutions to:

$\frac{-6}{x^2} + \frac{1}{x} + 1 = 0$

Possible Answers:

Correct answer:

Explanation:

Multiply both sides of the equation by x^2 , to get

-6 + x + x^2 = 0

This can be factored into the form

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

So we must solve

x+3 = 0

and

x-2 = 0

to get the solutions.

The solutions are:

x =-3, 2

and their sum is .

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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 #44 : Quadratic Equations And Inequalities

Example Question #45 : Quadratic Equations And Inequalities

Example Question #46 : Quadratic Equations And Inequalities

Example Question #81 : Intermediate Single Variable Algebra

Example Question #82 : Intermediate Single Variable Algebra

Example Question #83 : Intermediate Single Variable Algebra

Example Question #84 : Intermediate Single Variable Algebra

Example Question #23 : Finding Roots

Example Question #332 : Algebra Ii

Example Question #333 : Algebra Ii

All High School Math Resources

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