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High School Math : Quadratic Equations and Inequalities

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Example Questions

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Example Question #21 : 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 #1626 : High School Math

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 #1627 : High School Math

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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Example Question #1 : Completing The Square

$f(x)=x^{2}-4x+7$

Find the vertex of the parabola by completing the square.

Possible Answers:

$\small (2,3)$

$\small (3,2)$

$\small (1,4)$

$\small (7,0)$

$\small (0,7)$

Correct answer:

$\small (2,3)$

Explanation:

To find the vertex of a parabola, we must put the equation into the vertex form:

$\small f(x)=a[b(x-h)]^{2}+k$

The vertex can then be found with the coordinates (h, k).

To put the parabola's equation into vertex form, you have to complete the square. Completing the square just means adding the same number to both sides of the equation -- which, remember, doesn't change the value of the equation -- in order to create a perfect square.

Start with the original equation:

$f(x)=x^{2}-4x+7$

Put all of the terms on one side:

$f(x)-7=x^{2}-4x$

Now we know that we have to add something to both sides in order to create a perfect square:

$f(x)+7+\square =x^{2}-4x+\square$

In this case, we need to add 4 on both sides so that the right-hand side of the equation factors neatly.

$f(x)-7+4=x^{2}-4x+4$

Now we factor:

$f(x)-3=(x-2)^{2}$

Once we isolate f(x) , we have the equation in vertex form:

$f(x)=(x-2)^{2}+3$

Thus, the parabola's vertex can be found at $\small (2,3)$ .

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Example Question #2 : Completing The Square

Complete the square:

2x^2+3x+2=0

Possible Answers:

$x=\frac{-2 \pm i\sqrt{7}}{3}$

$x=\frac{-3 \pm i\sqrt{7}}{3}$

$x=\frac{-2 \pm i\sqrt{7}}{4}$

$x=\frac{-3 \pm i\sqrt{7}}{4}$

$x=\frac{-1 \pm i\sqrt{7}}{4}$

Correct answer:

$x=\frac{-3 \pm i\sqrt{7}}{4}$

Explanation:

Begin by dividing the equation by and subtracting from each side:

2x^2+3x+2=0

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

Square the value in front of the and add to each side:

$x^2+\frac{3}{2}x +\frac{9}{16} = -1+\frac{9}{16}$

Factor the left side of the equation:

$(x+\frac{3}{4})^2 = \frac{-7}{16}$

Take the square root of both sides and simplify:

$x+\frac{3}{4} = \frac{\pm \sqrt{-7}}{4}$

$x = \frac{-3 \pm i\sqrt{7}}{4}$

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Example Question #3 : Completing The Square

Use factoring to solve the quadratic equation:

6x^2+7x-3=0

Possible Answers:

$x=-\frac{3}{2},\frac{1}{4}$

$x=-\frac{3}{4},\frac{1}{4}$

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

$x=-\frac{3}{4},\frac{1}{3}$

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

Correct answer:

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

Explanation:

Factor and solve:

6x^2+7x-3=0

6x^2 +9x-2x-3=0

Factor like terms:

3x(2x+3)-1(2x+3)=0

Combine like terms:

(2x+3)(3x-1)=0

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

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Example Question #1 : Completing The Square

Complete the square:

2n^2+n-21=0

Possible Answers:

$n=-\frac{7}{2},3$

$n=-\frac{5}{2},3$

$n=-\frac{9}{2},3$

$n=-\frac{7}{2},4$

$n=-\frac{7}{2},2$

Correct answer:

$n=-\frac{7}{2},3$

Explanation:

Begin by dividing the equation by and adding 10.5 to each side:

2n^2+n-21=0

$n^2+\frac{1}{2}n = \frac{21}{2}$

Square the value in front of the and add to each side:

$n^2+\frac{1}{2}n + \frac{1}{16} = \frac{21}{2} + \frac{1}{16}$

Factor the left side of the equation:

$(n+\frac{1}{4})^2 = \frac{169}{16}$

Take the square root of both sides and simplify:

$n+\frac{1}{4} = \frac{\pm 13}{4}$

$n = \frac{-1 \pm 13}{4}$

$n=-\frac{7}{2},3$

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Example Question #5 : Completing The Square

Complete the square:

3x^2+4x+2=0

Possible Answers:

$x=\frac{-3 \pm i\sqrt{2}}{3}$

$x=\frac{-4 \pm i\sqrt{2}}{3}$

$x=\frac{-2 \pm i\sqrt{2}}{4}$

$x=\frac{-2 \pm i\sqrt{2}}{3}$

$x=\frac{-2 \pm i\sqrt{3}}{3}$

Correct answer:

$x=\frac{-2 \pm i\sqrt{2}}{3}$

Explanation:

Begin by dividing the equation by and subtracting $\frac{2}{3}$ from each side:

3x^2+4x+2=0

$x^2+\frac{4}{3}x = -\frac{2}{3}$

Square the value in front of the and add to each side:

$x^2+\frac{4}{3}x +\frac{4}{9} = -\frac{2}{3}+\frac{4}{9}$

Factor the left side of the equation:

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

Take the square root of both sides and simplify:

$x+\frac{2}{3} = \frac{\pm \sqrt{-2}}{3}$

$x = \frac{-2 \pm i\sqrt{2}}{3}$

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Example Question #91 : Intermediate Single Variable Algebra

Solve using the quadratic formula:

2x^2+2x+3=0

Possible Answers:

$x= \frac{-2 \pm i\sqrt{5}}{2}$

$x= \frac{-1 \pm i\sqrt{5}}{2}$

$x= \frac{-1 \pm i\sqrt{5}}{3}$

$x= \frac{-3 \pm i\sqrt{5}}{2}$

$x= \frac{-2 \pm i\sqrt{5}}{3}$

Correct answer:

$x= \frac{-1 \pm i\sqrt{5}}{2}$

Explanation:

Use the quadratic formula to solve:

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

a = 2

b = 2

c = 3

$x = \frac{-(2) \pm \sqrt{(2)^2-4(2)(3)}}{2(2)}$

$x = \frac{-2 \pm \sqrt{-20}}{4}$

$x = \frac{-2 \pm 2i\sqrt{5}}{4}$

$x = \frac{-1 \pm i\sqrt{5}}{2}$

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Example Question #51 : Quadratic Equations And Inequalities

Solve using the quadratic formula:

2a^2-5a+4=0

Possible Answers:

$a=\frac{5 \pm i\sqrt{7}}{4}$

$a=\frac{7 \pm i\sqrt{7}}{4}$

$a=\frac{4 \pm i\sqrt{7}}{4}$

$a=\frac{6 \pm i\sqrt{7}}{4}$

$a=\frac{3 \pm i\sqrt{7}}{4}$

Correct answer:

$a=\frac{5 \pm i\sqrt{7}}{4}$

Explanation:

Use the quadratic formula to solve:

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

a = 2

b = -5

c = 4

$a = \frac{-(-5) \pm \sqrt{(-5)^2-4(2)(4)}}{2(2)}$

$a = \frac{5 \pm \sqrt{-7}}{4}$

$a = \frac{5 \pm i\sqrt{7}}{4}$

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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 #21 : Finding Roots

Example Question #1626 : High School Math

Example Question #1627 : High School Math

Example Question #1 : Completing The Square

Example Question #2 : Completing The Square

Example Question #3 : Completing The Square

Example Question #1 : Completing The Square

Example Question #5 : Completing The Square

Example Question #91 : Intermediate Single Variable Algebra

Example Question #51 : Quadratic Equations And Inequalities

All High School Math Resources

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