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Algebra II : Quadratic Equations and Inequalities

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Example Questions

Example Question #241 : Quadratic Equations And Inequalities

2x^2-5x=8

In the above equation, what must next be done to both sides of the equation when completing the square?

Possible Answers:

Subtract 5x.

Divide by 2.

Add 5x.

Divide by -5.

Divide by 5.

Correct answer:

Divide by 2.

Explanation:

When completing the square, the lead coefficient should be one. To achieve this, divide both sides of the equation by the coefficient of the squared term.

$(2x^2-5x)\div2=x^2-\frac{5}{2}x$

and

$8\div2=4$

This leaves you with the equation $x^2-\frac{5}{2}x=4$

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Example Question #16 : Completing The Square

$x^2-\frac{5}{2}x=4$

In the above equation, what should be added to both sides of the equation in order to complete the square?

Possible Answers:

$-\frac{5}{4}$

$\frac{5}{2}$

$\frac{5}{4}$

$-\frac{25}{16}$

$\frac{25}{16}$

Correct answer:

$\frac{25}{16}$

Explanation:

a. Once the variables are on the left, the constant is on the right, and the lead coefficient is 1, you will create a perfect square trinomial on the left side of the equation. Do this by starting with the coefficient of the x term.

$\frac{5}{2}$

b. Divide this by 2.

$\frac{5}{2}\div2=\frac{5}{4}$

c. Square this term.

$(\frac{5}{4})^2=\frac{25}{16}$

d. Add the result to both sides of the equation.

$x^2-\frac{5}{2}x+\frac{25}{16}=\frac{89}{16}$

e. The expression on the left side of the equation is now a perfect square trinomial and can be factored to:

$(x-\frac{5}{4})^2=\frac{89}{16}$

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Example Question #87 : Solving Quadratic Equations

$(x-\frac{5}{4})^2=\frac{89}{16}$

Finish solving the above equation by completing the square.

Possible Answers:

$\frac{47}{2}$

$x=\frac{5}{4}\pm\frac{\sqrt89}{4}$

$x=\frac{5}{4}+\frac{\sqrt89}{4}$

$x=\frac{5}{4}-\frac{\sqrt89}{4}$

Correct answer:

$x=\frac{5}{4}\pm\frac{\sqrt89}{4}$

Explanation:

a. Take the square root of both sides. Don't forget to account for both positive and negative answers.

$\sqrt{(x-\frac{5}{4})^2}=\pm\sqrt{\frac{89}{16}}$

b. Simplify both sides

$x-\frac{5}{4}=\pm\frac{\sqrt{89}}{4}$

c. Isolate the variable by adding 5/4 to both sides.

$x=\frac{5}{4}\pm\frac{\sqrt{89}}{4}$

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Example Question #242 : Quadratic Equations And Inequalities

3x^2+12x=-5

Using the above equation, what should the next step look like when completing the square?

Possible Answers:

x^2+4x=-5

3x^2=-12x-5

$x^2+4x=-\frac{5}{3}$

x^2+12x=-5

Correct answer:

$x^2+4x=-\frac{5}{3}$

Explanation:

a. Once the variables are on the left side of the equation and the constant is on the right, make the lead coefficient 1 by dividing both sides of the equation by the coefficient of the squared term.

$(3x^2+12x)\div3=-5\div3$

b. Simplify

$x^2+4x=-\frac{5}{3}$

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Example Question #89 : Solving Quadratic Equations

$x^2+4x=-\frac{5}{3}$

In the above equation, what must next be done to both sides of the equation when completing the square?

Possible Answers:

Add 4.

Divide by 4.

Subtract 4.

Add 4x.

Correct answer:

Add 4.

Explanation:

To complete the square, you must create a perfect square trinomial on the left side of the equation.

Do this by starting with the coefficient of the x term.

b. Divide this by 2.

$4\div2=2$

c. Square this term.

2^2=4

d. Add the result to both sides of the equation.

$x^2+4x+4=4-\frac{5}{3}$

e. The expression on the left side of the equation is now a perfect square trinomial and can be factored to:

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

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Example Question #90 : Solving Quadratic Equations

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

Finish solving the above equation by completing the square.

Possible Answers:

$x=2+\sqrt{\frac{7}{3}}$

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

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

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

Correct answer:

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

Explanation:

a. Take the square root of both sides. Don't forget to account for both positive and negative answers.

$\sqrt{(x+2)^2}=\pm\sqrt{\frac{7}{3}}$

b. Simplify both sides, if possible.

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

c. Isolate the variable by subtracting 2 from both sides.

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

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Example Question #91 : Solving Quadratic Equations

Solve $\small x^2-4x=4x+65$ by completing the square.

Possible Answers:

x=13

x=9

$x=4\ \textup{and}\ x=0$

$x=13\ \textup{and}\ x=-5$

x=4

Correct answer:

$x=13\ \textup{and}\ x=-5$

Explanation:

To complete the square, we need to have the x terms on one side and the numbers on the other. Therefore,

$\small \small x^2 -4x=4x+65$

becomes

$\small \small x^2-8x=65$

When we want to complete the square, we want an equation in the form $\small \small \small a^2 + 2ab + b^2$ or $\small \small a^2 -2ab+b^2$ so that we can factor it into $\small \small (a+ b)^{2}$ or $\small (a-b)^2$ . To do this, we take half of the numeric portion of what we want our b term to be (in this problem $\small \small -8x$ ) and square it, therefore:

$\small b^2 = (\frac{1}{2}(-8))^{2}=(-4)^2=16$

Therefore, we add 16 to each side to obtain:

$\small x^2-8x+16 = 81$

$\small (x-4)^{2}=81$

$\small x-4=\pm9$

$\small x=13$ and $\small x=-5$

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Example Question #92 : Solving Quadratic Equations

Solve $\small \small \small 3x^2 + 36x=-33$ by completing the square.

Possible Answers:

$\small x=-6$

$\small x=\pm5$

$x=-6\ \textup{and}\ x=0$

$\small x=-1$

$x=-1\ \textup{and}\ x=-11$

Correct answer:

$x=-1\ \textup{and}\ x=-11$

Explanation:

To complete the square, we need the left side in a form $\small a^2 + 2ab +b^2$ or $\small a^2-2ab+b^2$ so that we can factor it into form $\small (a+b)^2$ or $\small (a-b)^2$ .

To do this, we first divide out three on the left-hand side to obtain:

$\small 3(x^2+12x)=-33$

We then take 1/2 of the number in our $\small 2ab$ term (in this case $\small 12x$ ) to obtain $\small b^2$ :

$\small b^2=(\frac{1}{2}(12))^2=(6)^2=36$

We then must add this to each side, but because we are completing the square inside of a parenthesis which is being multiplied by 3, we don't add 36 to each side, but rather 3 times 36, or 108. Therefore, we obtain:

$\small 3(x^2 +12x+36)=-33+108$

$\small 3(x+6)^2=75$

$\small (x+6)^{2}=25$

$\small x+6=\pm5$

$\small x=-1$ and $\small x=-11$

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Example Question #411 : Intermediate Single Variable Algebra

Solve for .

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

Possible Answers:

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

$x=2\pm\sqrt5$

$x=4\pm2$

$\textup{No solution}$

$x=-2+\sqrt{5}$

Correct answer:

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

Explanation:

Once the polynominal is factored out and everything is moved to the left, the equation becomes x^2+4x-1=0 which does not factor evenly, so you could use the quadratic formula, or complete the square. To complete the square, a coefficient must be found to factor the polynominal into a perfect square. The polynominal (x-a)^2 factors to x^2+2a+a^2 so we know that 2a=4 so a=2 and a^2=4 . To complete the square you add and subtract a^2 from the left side of the equation and strategically place parentheses to get (x^2+4x+4)-4-1=0 , this simplifies to (x+2)^2=5 , which simplifies to $x=-2\pm{\sqrt{5}}$ ,

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Example Question #411 : Intermediate Single Variable Algebra

What is/are the solution(s) to the quadratic equation

x^2 + 4x = 5 .

Hint: Complete the square

Possible Answers:

x=3

x=1, -5

x=2,4

x=1,5

Correct answer:

x=1, -5

Explanation:

x^2 + 4x = 5

When using the complete the square method we will divide the coefficient by two and then square it. This will become our term which we will add to both sides.

In the form,

ax^2+bx+c

our a=1 , b=4 and we will complete the square to find the value.

Therefore we get:

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

x^2 + 4x +4= 9

(x+2)^2= 9

$x+2=\pm3$

x=1,-5

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Algebra II : Quadratic Equations and Inequalities

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #241 : Quadratic Equations And Inequalities

Example Question #16 : Completing The Square

Example Question #87 : Solving Quadratic Equations

Example Question #242 : Quadratic Equations And Inequalities

Example Question #89 : Solving Quadratic Equations

Example Question #90 : Solving Quadratic Equations

Example Question #91 : Solving Quadratic Equations

Example Question #92 : Solving Quadratic Equations

Example Question #411 : Intermediate Single Variable Algebra

Example Question #411 : Intermediate Single Variable Algebra

All Algebra II Resources

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