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High School Math : High School Math

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8 Diagnostic Tests 613 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

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 #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 #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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Example Question #1 : Completing The Square

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

Find the vertex of the parabola by completing the square.

Possible Answers:

$\small (7,0)$

$\small (2,3)$

$\small (1,4)$

$\small (3,2)$

$\small (0,7)$

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 #31 : Solving Quadratic Equations

Complete the square:

2x^2+3x+2=0

Possible Answers:

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

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

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

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

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

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:

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

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

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

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

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

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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High School Math : High School Math

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

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 #21 : Finding Roots

Example Question #332 : Algebra Ii

Example Question #333 : Algebra Ii

Example Question #1 : Completing The Square

Example Question #31 : Solving Quadratic Equations

Example Question #3 : Completing The Square

All High School Math Resources

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