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Algebra II : Intermediate Single-Variable Algebra

Study concepts, example questions & explanations for Algebra II

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Example Questions

Example Question #1561 : Algebra Ii

Solve for x by completing the square:

2x^2+10x-12=0

Possible Answers:

No solution

$x=\sqrt{13}$

$x=\pm \frac{\sqrt{37}}{2}-\frac{5}{2}$

x=0

Correct answer:

$x=\pm \frac{\sqrt{37}}{2}-\frac{5}{2}$

Explanation:

To solve for x by completing the square, first we must leave only terms containing x on one side:

2x^2+10x=6

Now, we can divide by 2 on both sides so that we have a coefficient of 1 for the leading term (this isn't necessary but makes the process easier):

x^2+5x=3

Now, on the left side, we must complete the square. We are trying to create

x^2+2ax+a^2=(x+a)^2

so we must halve the coefficient of x, and then square it. When we add this term to the left side, we must also add it to the right side of the equation (so we aren't changing the function):

$x^2+5x+\frac{25}{4}=3+\frac{25}{4}$

Now, rewrite the left hand side as a perfect square:

$(x+\frac{5}{2})^2=\frac{37}{4}$

Now, we can solve for x:

$x+\frac{5}{2}=\pm\sqrt{\frac{37}{4}}$

$x=\pm \frac{\sqrt{37}}{2}-\frac{5}{2}$

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Example Question #43 : Completing The Square

Solve: x^2+4x-32=0

Possible Answers:

$x=2\pm-6$

$x=2\pm6$

$x=-2\pm4$

x=4,-8

None of these.

Correct answer:

x=4,-8

Explanation:

x^2+4x-32=0

To solve for x factor the quadratic into two binomials. To factor the quadratic, look at the factors for the constant value.

Since the middle value is positive and the last term is negative, this signifies that one binomial will contain a negative sign and the other will be positive.

$(x- \ )(x+\ )$

Now let's look at the factors of the third term.

32: [1,32], [2,16],[4,8]

From here, which set of factors when added together result in the value of the middle term, or in this case .

(x- 4 )(x+8)

Since

(-4)+8=4 .

Now set each binomial equal to zero and solve for x.

Binomial 1:

x-4=0

____________

x=4

Binomial 2:

x+8=0

_____________

x=-8

Therefore the solution is,

x=4,-8

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Example Question #431 : Intermediate Single Variable Algebra

Which of the following equations is equivalent to x^2+4x=-7 after completing the square?

Possible Answers:

(x+2)^2=-28

(x+2)^2=-3

(x+2)^2=-12

(x+2)^2=-11

(x+2)^2=7

Correct answer:

(x+2)^2=-3

Explanation:

In order to complete the square, divide the coefficient in the equation x^2+4x=-7 by two and square the quantity.

$(\frac{4}{2})^2 = (2)^2=4$

This value must be added to both sides of the equation.

x^2+4x+4=-7+4

Factorize the left side and simplify the right side.

The answer is: (x+2)^2=-3

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Example Question #271 : Quadratic Equations And Inequalities

What will be the equation after completing the square? x^2-4x = -7

Possible Answers:

(x-2)^2= -4

(x-2)^2= -3

(x+2)^2= -3

(x-2)^2= -10

(x+2)^2=- 4

Correct answer:

(x-2)^2= -3

Explanation:

To complete the square, we will need to divide the coefficient of the x-variable on the left side of the equation.

$-4x \rightarrow\frac{-4}{2} = 2$

Square this number, and add it on both sides of the equation.

x^2-4x+4 = -7+4

Notice that the left side of the equation can be factorized by a single binomial squared. Simplify the right side as well.

The answer is: (x-2)^2= -3

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Example Question #46 : Completing The Square

Solve by completing the square:

x^2-6x+2=0

Possible Answers:

$x=4.12\text{ or }0.33$

$x=9.12\text{ or }0.13$

$x=5.65\text{ or }0.35$

$x=3.29 \text{ or }1.31$

Correct answer:

$x=5.65\text{ or }0.35$

Explanation:

Start by moving the number to the right of the equation so that all the terms with values are alone:

x^2-6x+2=0

x^2-6x=-2

Now, you need to figure out what number to add to both sides. To do so, take the coefficient in front of the term, divide it by , then square it:

Coefficient:

$\text{Term to add}=(\frac{-6}{2})^2=(-3)^2=9$

Add this number to both sides of the equation:

x^2-6x+9=-2+9=7

Simplify the left side of the equation.

(x-3)^2=7

Now solve for .

$x-3=\pm\sqrt7$

$x=3+\sqrt7 = 5.65$ , or

$x=3-\sqrt7=0.35$

Make sure to round to places after the decimal.

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Example Question #47 : Completing The Square

Solve the following quadratic equation by completing the square:

x^2+14x-20=0

Possible Answers:

$x=0.19\text{ or }-13.22$

$x=3.65\text{ or }-10.01$

$x=5.61\text{ or }-14.19$

$x=1.17\text{ or }-20.51$

$x=1.31\text{ or }-15.31$

Correct answer:

$x=1.31\text{ or }-15.31$

Explanation:

Start by moving the number to the right of the equation so that all the terms with values are alone:

x^2+14x-20=0

x^2+14x=20

Now, you need to figure out what number to add to both sides. To do so, take the coefficient in front of the term, divide it by , then square it:

Coefficient:

$\text{Term to add}=(\frac{14}{2})^2=(7)^2=49$

Add this number to both sides of the equation:

x^2+14x+49=20+49=69

Simplify the left side of the equation.

(x+7)^2=69

Now solve for .

$x+7=\pm\sqrt{69}$

$x=-7+\sqrt{69}=1.31$ , or

$x=-7-\sqrt{69}=-15.31$

Make sure to round to places after the decimal.

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Example Question #48 : Completing The Square

Solve the following quadratic equation by completing the square:

x^2+12x-1=0

Possible Answers:

$x=1.19\text{ or }-10.68$

$x=0.12\text{ or }-14.12$

$x=0.09\text{ or }-19.47$

$x=0.08\text{ or }-12.08$

Correct answer:

$x=0.08\text{ or }-12.08$

Explanation:

Start by moving the number to the right of the equation so that all the terms with values are alone:

x^2+12x-1=0

x^2+12x=1

Now, you need to figure out what number to add to both sides. To do so, take the coefficient in front of the term, divide it by , then square it:

Coefficient:

$\text{Term to add}=(\frac{12}{2})^2=(6)^2=36$

Add this number to both sides of the equation:

x^2+12x+36=1+36=37

Simplify the left side of the equation.

(x+6)^2=37

Now solve for .

$x+6=\pm\sqrt{37}$

$x=-6+\sqrt{37}=0.08$ , or

$x=-6-\sqrt{37}=-12.08$

Make sure to round to places after the decimal.

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Example Question #49 : Completing The Square

Solve the following quadratic equation by completing the square:

2x^2+6x-9=0

Possible Answers:

$x=1.10\text{ or }-4.10$

$x=2.28\text{ or }-1.61$

$x=2.14\text{ or }-10.40$

$x=4.29\text{ or }-2.11$

Correct answer:

$x=1.10\text{ or }-4.10$

Explanation:

Start by dividing the entire equation by the coefficient in front of x^2 , which is .

2x^2+6x-9=0

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

Then, move the number to the right of the equation so that all the terms with values are alone:

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

Now, you need to figure out what number to add to both sides. To do so, take the coefficient in front of the term, divide it by , then square it:

Coefficient:

$\text{Term to add}=(\frac{3}{2})^2=\frac{9}{4}$

Add this number to both sides of the equation:

$x^2+3x+\frac{9}{4}=\frac{9}{2}+\frac{9}{4}=\frac{27}{4}$

Simplify the left side of the equation.

$(x+\frac{3}{2})^2=\frac{27}{4}$

Now solve for .

$x+\frac{3}{2}=\pm\sqrt{\frac{27}{4}}$

$x=-\frac{3}{2}+\sqrt{\frac{27}{4}}=1.10$ , or

$x=-\frac{3}{2}-\sqrt{\frac{27}{4}}=-4.10$

Make sure to round to places after the decimal.

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Example Question #50 : Completing The Square

Solve the quadratic equation by completing the square:

x^2-24x+100=0

Possible Answers:

$x=19.08\text{ or }4.11$

$x=17.76\text{ or }8.17$

$x=20.12\text{ or }12.55$

$x=18.63 \text{ or }5.37$

Correct answer:

$x=18.63 \text{ or }5.37$

Explanation:

Start by moving the number to the right of the equation so that all the terms with values are alone:

x^2-24x+100=0

x^2-24x=-100

Now, you need to figure out what number to add to both sides. To do so, take the coefficient in front of the term, divide it by , then square it:

Coefficient:

$\text{Term to add}=(\frac{-24}{2})^2=(-12)^2=144$

Add this number to both sides of the equation:

x^2-24x+144=-100+144=44

Simplify the left side of the equation.

(x-12)^2=44

Now solve for .

$x-12=\pm\sqrt{44}$

$x=12+\sqrt{44}=18.63$ , or

$x=12-\sqrt{44}=5.37$

Make sure to round to places after the decimal.

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Example Question #51 : Completing The Square

Solve the following quadratic equation by completing the square:

x^2-10x+9=0

Possible Answers:

$x=4\text{ or }-9$

$x=-2\text{ or }2$

$x=1\text{ or } 9$

$x=-1\text{ or }-9$

Correct answer:

$x=1\text{ or } 9$

Explanation:

Start by moving the number to the right of the equation so that all the terms with values are alone:

x^2-10x+9=0

x^2-10x=-9

Now, you need to figure out what number to add to both sides. To do so, take the coefficient in front of the term, divide it by , then square it:

Coefficient:

$\text{Term to add}=(\frac{10}{2})^2=(5)^2=25$

Add this number to both sides of the equation:

x^2-10x+25=-9+25=16

Simplify the left side of the equation.

(x-5)^2=16

Now solve for .

$x-5=\pm4$

x=5+4=9 , or

x=5-4=1

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Algebra II : Intermediate Single-Variable Algebra

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #1561 : Algebra Ii

Example Question #43 : Completing The Square

Example Question #431 : Intermediate Single Variable Algebra

Example Question #271 : Quadratic Equations And Inequalities

Example Question #46 : Completing The Square

Example Question #47 : Completing The Square

Example Question #48 : Completing The Square

Example Question #49 : Completing The Square

Example Question #50 : Completing The Square

Example Question #51 : Completing The Square

All Algebra II Resources

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