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Algebra II : Solving Quadratic Equations

Study concepts, example questions & explanations for Algebra II

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Example Questions

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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Example Question #121 : Solving Quadratic Equations

Solve the quadratic equation by completing the square:

5x^2+20x-12=0

Possible Answers:

$x=0.53\text{ or }-4.53$

$x=1.01\text{ or }-2.15$

$x=0.09\text{ or }-1.66$

$x=0.71\text{ or }1.54$

Correct answer:

$x=0.53\text{ or }-4.53$

Explanation:

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

5x^2+20x-12=0

$x^2+4x-\frac{12}{5}=0$

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

$x^2+4x=\frac{12}{5}$

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:

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

Add this number to both sides of the equation:

$x^2+4x+4=\frac{12}{5}+4=\frac{32}{5}$

Simplify the left side of the equation.

$(x+2)^2=\frac{32}{5}$

Now solve for .

$x+2=\pm\sqrt{\frac{32}{5}}$

$x=-2+\sqrt{\frac{32}{5}}=0.53$ , or

$x=-2-\sqrt{\frac{32}{5}}=-4.53$

Make sure to round to places after the decimal.

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Example Question #122 : Solving Quadratic Equations

Solve the quadratic equation by completing the square:

x^2-18x+46=0

Possible Answers:

$x=20.59\text{ or }4.12$

$x=13.44\text{ or }6.24$

$x=14.92\text{ or }3.08$

$x=19.90\text{ or }0.18$

Correct answer:

$x=14.92\text{ or }3.08$

Explanation:

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

x^2-18x+46=0

x^2-18x=-46

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:

$\text{Term to add}=(\frac{-18}{2})^2=(-9)^2=81$

Add this number to both sides of the equation:

x^2-18x+81=-46+81=35

Simplify the left side of the equation.

(x-9)^2=35

Now solve for .

$x-9=\pm\sqrt{35}$

$x=9+\sqrt{35}=14.92$ , or

$x=9-\sqrt{35}=3.08$

Make sure to round to places after the decimal.

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Example Question #123 : Solving Quadratic Equations

Solve the quadratic equation by completing the square:

x^2+22x+100=0

Possible Answers:

$x=-9.01\text{ or }-18.23$

$x=-9.29\text{ or }-21.32$

$x=-5.45\text{ or }-15.61$

$x=-6.42\text{ or }-15.58$

Correct answer:

$x=-6.42\text{ or }-15.58$

Explanation:

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

x^2+22x+100=0

x^2+22x=-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:

$\text{Term to add}=(\frac{22}{2})^2=(11)^2=121$

Add this number to both sides of the equation:

x^2+22x+121=-100+121=21

Simplify the left side of the equation.

(x+11)^2=21

Now solve for .

$x+11=\pm\sqrt{21}$

$x=-11+\sqrt{21}=-6.42$ , or

$x=-11-\sqrt{21}=-15.58$

Make sure to round to places after the decimal.

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Example Question #55 : Completing The Square

Solve the quadratic equation by completing the square:

x^2-30x+98=0

Possible Answers:

$x=21.92\text{ or }12.28$

$x=26.27\text{ or }3.73$

$x=29.71\text{ or }9.22$

$x=25.67\text{ or }8.21$

Correct answer:

$x=26.27\text{ or }3.73$

Explanation:

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

x^2-30x+98=0

x^2-30x=-98

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:

$\text{Term to add}=(\frac{30}{2})^2=(15)^2=225$

Add this number to both sides of the equation:

x^2-30x+225=-98+225=127

Simplify the left side of the equation.

(x-15)^2=127

Now solve for .

$x-15=\pm\sqrt{127}$

$x=15+\sqrt{127}=26.27$ , or

$x=15-\sqrt{127}=3.73$

Make sure to round to places after the decimal.

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Example Question #56 : Completing The Square

Solve the following quadratic equation by completing the square:

x^2-16x+40=0

Possible Answers:

$x=-5.19\text{ or }10.72$

$x=4.19\text{ or }19.52$

$x=-3.09\text{ or }14.98$

$x=3.10\text{ or }12.90$

Correct answer:

$x=3.10\text{ or }12.90$

Explanation:

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

x^2-16x+40=0

x^2-16x=-40

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:

$\text{Term to add}=(\frac{-16}{2})^2=(-8)^2=64$

Add this number to both sides of the equation:

x^2-16x+64=-40+64=24

Simplify the left side of the equation.

(x-8)^2=24

Now solve for .

$x-8=\pm\sqrt{24}$

$x=8+\sqrt{24}=12.90$ , or

$x=8-\sqrt{24}=3.10$

Make sure to round to places after the decimal.

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Example Question #1581 : Algebra Ii

Which of the following is the same after completing the square?

3x^2+x-6 =0

Possible Answers:

$(x+\frac{1}{6})^2 =\frac{73}{36}$

$(x+\frac{1}{3})^2 =\frac{7}{2}$

$(x-\frac{1}{6})^2 =\frac{73}{36}$

$(x-\frac{1}{3})^2 =\frac{7}{2}$

$(x+\frac{1}{6})^2 =\frac{1}{2}$

Correct answer:

$(x+\frac{1}{6})^2 =\frac{73}{36}$

Explanation:

Divide by three on both sides.

$\frac{3x^2+x-6 }{3}=\frac{0}{3}$

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

Add two on both sides.

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

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

To complete the square, we will need to divide the one-third coefficient by two, which is similar to multiplying by one half, square the quantity, and add the two values on both sides.

$x^2+\frac{1}{3}x+ (\frac{1}{3}\cdot \frac{1}{2})^2 = 2+ (\frac{1}{3}\cdot \frac{1}{2})^2$

Simplify both sides.

$x^2+\frac{1}{3}x+ \frac{1}{36} = 2+ \frac{1}{36}$

Factor the left side, and combine the terms on the right.

$(x+\frac{1}{6})^2 = \frac{72}{36}+ \frac{1}{36}$

The answer is: $(x+\frac{1}{6})^2 =\frac{73}{36}$

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Example Question #1581 : Algebra Ii

Solve for by completing the square.

x^2+8x-2=0

Possible Answers:

$x=0.39\text{ and }x=-4.39$

$x=1.24\text{ and }x=-10.24$

$x=0.24\text{ and }x=-8.24$

$x=0.97\text{ and }x=-8.97$

Correct answer:

$x=0.24\text{ and }x=-8.24$

Explanation:

x^2+8x-2=0

Start by adding to both sides so that the terms with the are together on the left side of the equation.

x^2+8x=2

Now, look at the coefficient of the -term. To complete the square, divide this coefficient by , then square the result. Add this term to both sides of the equation.

x^2+8x+16=2+16

Rewrite the left side of the equation in the squared form.

(x+4)^2=18

Take the square root of both sides.

$x+4=\pm \sqrt{18}$

Now solve for .

$x=-4+\sqrt{18}=0.24$

$x=-4-\sqrt{18}=-8.24$

Round to two places after the decimal.

Report an Error

Example Question #1582 : Algebra Ii

Solve for by completing the square.

x^2-6x+2=0

Possible Answers:

$x=5.65\text{ and }x=0.35$

$x=1.95\text{ and }x=5.05$

$x=-8.75\text{ and }x=-2.25$

$x=10.55\text{ and }x=2.45$

Correct answer:

$x=5.65\text{ and }x=0.35$

Explanation:

x^2-6x+2=0

Start by subtracting from both sides so that the terms with the are together on the left side of the equation.

x^2-6x=-2

Now, look at the coefficient of the -term. To complete the square, divide this coefficient by , then square the result. Add this term to both sides of the equation.

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

Rewrite the left side of the equation in the squared form.

(x-3)^2=7

Take the square root of both sides.

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

Now solve for .

$x=3+\sqrt{7}=5.65$

$x=3-\sqrt{7}=0.35$

Round to two places after the decimal.

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Example Question #1582 : Algebra Ii

Solve for by completing the square.

x^2-10x+20=0

Possible Answers:

$x=3.33\text{ and }x=8.67$

$x=-5.21\text{ and }x=-9.79$

$x=8.06\text{ and }x=1.94$

$x=7.24\text{ and }x=2.76$

Correct answer:

$x=7.24\text{ and }x=2.76$

Explanation:

x^2-10x+20=0

Start by subtracting from both sides so that the terms with the are together on the left side of the equation.

x^2-10x=-20

Now, look at the coefficient of the -term. To complete the square, divide this coefficient by , then square the result. Add this term to both sides of the equation.

x^2-10x+25=-20+25

Rewrite the left side of the equation in the squared form.

(x-5)^2=5

Take the square root of both sides.

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

Now solve for .

$x=5+\sqrt{5}=7.24$

$x=5-\sqrt{5}=2.76$

Round to two places after the decimal.

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Algebra II : Solving Quadratic Equations

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #51 : Completing The Square

Example Question #121 : Solving Quadratic Equations

Example Question #122 : Solving Quadratic Equations

Example Question #123 : Solving Quadratic Equations

Example Question #55 : Completing The Square

Example Question #56 : Completing The Square

Example Question #1581 : Algebra Ii

Example Question #1581 : Algebra Ii

Example Question #1582 : Algebra Ii

Example Question #1582 : Algebra Ii

All Algebra II Resources

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