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Algebra II : Quadratic Equations and Inequalities

Study concepts, example questions & explanations for Algebra II

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Example Questions

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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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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Algebra II : Quadratic Equations and Inequalities

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

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

Example Question #121 : Solving Quadratic Equations

Example Question #122 : Solving Quadratic Equations

Example Question #123 : Solving Quadratic Equations

All Algebra II Resources

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