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Common Core: High School - Algebra : High School: Algebra

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All Common Core: High School - Algebra Resources

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Example Questions

Example Question #501 : High School: Algebra

Solve $x^{2} + 41 x - 19 = 0$ by completing the square. Round your answer to the nearest hundredth.

Possible Answers:

x = -0.04, x = -40.96

x = -2.03, x = 4.45

x = 0.46 , x = -41.46

x = -2.04, x = -38.96

x = 19.33, x = -7.84

Correct answer:

x = 0.46 , x = -41.46

Explanation:

The first step is to add to both sides.

$x^{2} + 41 x - 19 + 19 = 0 + 19$

$x^{2} + 41 x = 19$

Now we take the coefficient in front of the $\uptext{x}$ term, divide it by , square it and add it to each side.

$x^{2} + 41 x + \frac{41}{2} ^2 = 19 + \frac{41}{2} ^2$

$x^{2} + 41 x + \frac{1681}{4} = 19 + \frac{1681}{4}$

Now we factor the left hand side, and add up the right hand side.

$\left( x + \frac{41}{2} \right)^2= \frac{1757}{4}$

Now we take the square root of each side.

$\sqrt{\left( x + \frac{41}{2} \right)^2}= \sqrt{ \frac{1757}{4} }$

$x + \frac{41}{2} = \sqrt{ \frac{1757}{4} }$

Now we subtract $\frac{41}{2}$ from each side.

$x + \frac{41}{2} - \frac{41}{2} = \sqrt{ \frac{1757}{4} }- \frac{41}{2}$

$x = - \frac{41}{2} + \frac{\sqrt{1757}}{2}$

Since we are taking the square root, we need to set up equations to solve for $\uptext{x}$ .

$x = - \frac{41}{2} + \frac{\sqrt{1757}}{2}$

x = 0.46

$x = - \frac{\sqrt{1757}}{2} - \frac{41}{2}$

x = -41.46

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Example Question #502 : High School: Algebra

Solve $x^{2} + 57 x - 87 = 0$ by completing the square. Round your answer to the nearest hundredth.

Possible Answers:

x = 0.99, x = -57.99

x = -1.01, x = -55.99

x = 1.49 , x = -58.49

x = 4.16, x = -6.5

x = -27.31, x = -2.9

Correct answer:

x = 1.49 , x = -58.49

Explanation:

The first step is to add to both sides.

$x^{2} + 57 x - 87 + 87 = 0 + 87$

$x^{2} + 57 x = 87$

Now we take the coefficient in front of the $\uptext{x}$ term, divide it by , square it and add it to each side.

$x^{2} + 57 x + \frac{57}{2} ^2 = 87 + \frac{57}{2} ^2$

$x^{2} + 57 x + \frac{3249}{4} = 87 + \frac{3249}{4}$

Now we factor the left hand side, and add up the right hand side.

$\left( x + \frac{57}{2} \right)^2= \frac{3597}{4}$

Now we take the square root of each side.

$\sqrt{\left( x + \frac{57}{2} \right)^2}= \sqrt{ \frac{3597}{4} }$

$x + \frac{57}{2} = \sqrt{ \frac{3597}{4} }$

Now we subtract $\frac{57}{2}$ from each side.

$x + \frac{57}{2} - \frac{57}{2} = \sqrt{ \frac{3597}{4} }- \frac{57}{2}$

$x = - \frac{57}{2} + \frac{\sqrt{3597}}{2}$

Since we are taking the square root, we need to set up equations to solve for $\uptext{x}$ .

$x = - \frac{57}{2} + \frac{\sqrt{3597}}{2}$

x = 1.49

$x = - \frac{\sqrt{3597}}{2} - \frac{57}{2}$

x = -58.49

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Example Question #503 : High School: Algebra

Solve $x^{2} + 29 x - 45 = 0$ by completing the square. Round your answer to the nearest hundredth.

Possible Answers:

x = -1.02, x = -27.98

x = 3.69, x = -6.08

x = -29.41, x = 47.98

x = 0.98, x = -29.98

x = 1.48 , x = -30.48

Correct answer:

x = 1.48 , x = -30.48

Explanation:

The first step is to add to both sides.

$x^{2} + 29 x - 45 + 45 = 0 + 45$

$x^{2} + 29 x = 45$

Now we take the coefficient in front of the $\uptext{x}$ term, divide it by , square it and add it to each side.

$x^{2} + 29 x + \frac{29}{2} ^2 = 45 + \frac{29}{2} ^2$

$x^{2} + 29 x + \frac{841}{4} = 45 + \frac{841}{4}$

Now we factor the left hand side, and add up the right hand side.

$\left( x + \frac{29}{2} \right)^2= \frac{1021}{4}$

Now we take the square root of each side.

$\sqrt{\left( x + \frac{29}{2} \right)^2}= \sqrt{ \frac{1021}{4} }$

$x + \frac{29}{2} = \sqrt{ \frac{1021}{4} }$

Now we subtract $\frac{29}{2}$ from each side.

$x + \frac{29}{2} - \frac{29}{2} = \sqrt{ \frac{1021}{4} }- \frac{29}{2}$

$x = - \frac{29}{2} + \frac{\sqrt{1021}}{2}$

Since we are taking the square root, we need to set up equations to solve for $\uptext{x}$ .

$x = - \frac{29}{2} + \frac{\sqrt{1021}}{2}$

x = 1.48

$x = - \frac{\sqrt{1021}}{2} - \frac{29}{2}$

x = -30.48

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Example Question #504 : High School: Algebra

Solve $x^{2} + 25 x - 40 = 0$ by completing the square. Round your answer to the nearest hundredth.

Possible Answers:

x = 1.01, x = -26.01

x = -9.39, x = 8.59

x = 1.51 , x = -26.51

x = 14.02, x = 45.3

x = -0.99, x = -24.01

Correct answer:

x = 1.51 , x = -26.51

Explanation:

The first step is to add 40 to both sides.

$x^{2} + 25 x - 40 + 40 = 0 + 40$

$x^{2} + 25 x = 40$

Now we take the coefficient in front of the $\uptext{x}$ term, divide it by , square it and add it to each side.

$x^{2} + 25 x + \frac{25}{2} ^2 = 40 + \frac{25}{2} ^2$

$x^{2} + 25 x + \frac{625}{4} = 40 + \frac{625}{4}$

Now we factor the left hand side, and add up the right hand side.

$\left( x + \frac{25}{2} \right)^2= \frac{785}{4}$

Now we take the square root of each side.

$\sqrt{\left( x + \frac{25}{2} \right)^2}= \sqrt{ \frac{785}{4} }$

$x + \frac{25}{2} = \sqrt{ \frac{785}{4} }$

Now we subtract $\frac{25}{2}$ from each side.

$x + \frac{25}{2} - \frac{25}{2} = \sqrt{ \frac{785}{4} }- \frac{25}{2}$

$x = - \frac{25}{2} + \frac{\sqrt{785}}{2}$

Since we are taking the square root, we need to set up equations to solve for $\uptext{x}$ .

$x = - \frac{25}{2} + \frac{\sqrt{785}}{2}$

x = 1.51

$x = - \frac{\sqrt{785}}{2} - \frac{25}{2}$

x = -26.51

Report an Error

Example Question #505 : High School: Algebra

Solve $x^{2} + 19 x - 2 = 0$ by completing the square. Round your answer to the nearest hundredth.

Possible Answers:

x = -0.4, x = -18.6

x = 0.1 , x = -19.1

x = -7.01, x = 5.94

x = -2.4, x = -16.6

x = -6.66, x = -6.47

Correct answer:

x = 0.1 , x = -19.1

Explanation:

The first step is to add to both sides.

$x^{2} + 19 x - 2 + 2 = 0 + 2$

$x^{2} + 19 x = 2$

Now we take the coefficient in front of the $\uptext{x}$ term, divide it by , square it and add it to each side.

$x^{2} + 19 x + \frac{19}{2} ^2 = 2 + \frac{19}{2} ^2$

$x^{2} + 19 x + \frac{361}{4} = 2 + \frac{361}{4}$

Now we factor the left hand side, and add up the right hand side.

$\left( x + \frac{19}{2} \right)^2= \frac{369}{4}$

Now we take the square root of each side.

$\sqrt{\left( x + \frac{19}{2} \right)^2}= \sqrt{ \frac{369}{4} }$

$x + \frac{19}{2} = \sqrt{ \frac{369}{4} }$

Now we subtract $\frac{19}{2}$ from each side.

$x + \frac{19}{2} - \frac{19}{2} = \sqrt{ \frac{369}{4} }- \frac{19}{2}$

$x = - \frac{19}{2} + \frac{3 \sqrt{41}}{2}$

Since we are taking the square root, we need to set up equations to solve for $\uptext{x}$ .

$x = - \frac{19}{2} + \frac{3 \sqrt{41}}{2}$

x = 0.1

$x = - \frac{3 \sqrt{41}}{2} - \frac{19}{2}$

x = -19.1

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Example Question #506 : High School: Algebra

Solve $x^{2} + 43 x - 19 = 0$ by completing the square. Round your answer to the nearest hundredth.

Possible Answers:

x = 0.44 , x = -43.44

x = -0.06, x = -42.94

x = -2.06, x = -40.94

x = 7.69, x = -8.09

x = 19.15, x = 30.43

Correct answer:

x = 0.44 , x = -43.44

Explanation:

The first step is to add to both sides.

$x^{2} + 43 x - 19 + 19 = 0 + 19$

$x^{2} + 43 x = 19$

Now we take the coefficient in front of the $\uptext{x}$ term, divide it by , square it and add it to each side.

$x^{2} + 43 x + \frac{43}{2} ^2 = 19 + \frac{43}{2} ^2$

$x^{2} + 43 x + \frac{1849}{4} = 19 + \frac{1849}{4}$

Now we factor the left hand side, and add up the right hand side.

$\left( x + \frac{43}{2} \right)^2= \frac{1925}{4}$

Now we take the square root of each side.

$\sqrt{\left( x + \frac{43}{2} \right)^2}= \sqrt{ \frac{1925}{4} }$

$x + \frac{43}{2} = \sqrt{ \frac{1925}{4} }$

Now we subtract $\frac{43}{2}$ from each side.

$x + \frac{43}{2} - \frac{43}{2} = \sqrt{ \frac{1925}{4} }- \frac{43}{2}$

$x = - \frac{43}{2} + \frac{5 \sqrt{77}}{2}$

Since we are taking the square root, we need to set up equations to solve for $\uptext{x}$ .

$x = - \frac{43}{2} + \frac{5 \sqrt{77}}{2}$

x = 0.44

$x = - \frac{5 \sqrt{77}}{2} - \frac{43}{2}$

x = -43.44

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Example Question #507 : High School: Algebra

Solve $x^{2} + 19 x - 34 = 0$ by completing the square. Round your answer to the nearest hundredth.

Possible Answers:

x = 1.65 , x = -20.65

x = -8.53, x = 0.05

x = -0.85, x = -18.15

x = -33.8, x = -13.2

x = 1.15, x = -20.15

Correct answer:

x = 1.65 , x = -20.65

Explanation:

The first step is to add to both sides.

$x^{2} + 19 x - 34 + 34 = 0 + 34$

$x^{2} + 19 x = 34$

Now we take the coefficient in front of the $\uptext{x}$ term, divide it by , square it and add it to each side.

$x^{2} + 19 x + \frac{19}{2} ^2 = 34 + \frac{19}{2} ^2$

$x^{2} + 19 x + \frac{361}{4} = 34 + \frac{361}{4}$

Now we factor the left hand side, and add up the right hand side.

$\left( x + \frac{19}{2} \right)^2= \frac{497}{4}$

Now we take the square root of each side.

$\sqrt{\left( x + \frac{19}{2} \right)^2}= \sqrt{ \frac{497}{4} }$

$x + \frac{19}{2} = \sqrt{ \frac{497}{4} }$

Now we subtract $\frac{19}{2}$ from each side.

$x + \frac{19}{2} - \frac{19}{2} = \sqrt{ \frac{497}{4} }- \frac{19}{2}$

$x = - \frac{19}{2} + \frac{\sqrt{497}}{2}$

Since we are taking the square root, we need to set up equations to solve for $\uptext{x}$ .

$x = - \frac{19}{2} + \frac{\sqrt{497}}{2}$

x = 1.65

$x = - \frac{\sqrt{497}}{2} - \frac{19}{2}$

x = -20.65

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Example Question #501 : High School: Algebra

Solve the following by completing the square. Round your answer to the nearest hundredth:

$x^{2} + 19 x - 52 = 0$

Possible Answers:

x = 1.93, x = -20.93

x = -3.38, x = 7.28

x = -5.2, x = -7.4

x = 2.43 , x = -21.43

x = -0.06, x = -18.93

Correct answer:

x = 2.43 , x = -21.43

Explanation:

The first step is to add to both sides.

$x^{2} + 19 x - 52 + 52 = 0 + 52$

$x^{2} + 19 x = 52$

Now we take the coefficient in front of the $\uptext{x}$ term, divide it by , square it and add it to each side.

$x^{2} + 19 x + \frac{19}{2} ^2 = 52 + \frac{19}{2} ^2$

$x^{2} + 19 x + \frac{361}{4} = 52 + \frac{361}{4}$

Now we factor the left hand side, and add up the right hand side.

$\left( x + \frac{19}{2} \right)^2= \frac{569}{4}$

Now we take the square root of each side.

$\sqrt{\left( x + \frac{19}{2} \right)^2}= \sqrt{ \frac{569}{4} }$

$x + \frac{19}{2} = \sqrt{ \frac{569}{4} }$

Now we subtract $\frac{19}{2}$ from each side.

$x + \frac{19}{2} - \frac{19}{2} = \sqrt{ \frac{569}{4} }- \frac{19}{2}$

$x = - \frac{19}{2} + \frac{\sqrt{569}}{2}$

Since we are taking the square root, we need to set up equations to solve for $\uptext{x}$ .

$x = - \frac{19}{2} + \frac{\sqrt{569}}{2}$

x = 2.43

$x = - \frac{\sqrt{569}}{2} - \frac{19}{2}$

x = -21.43

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Example Question #1 : Solve Quadratic Equations By Inspection, Quadratic Formula, Factoring, Completing The Square, And Taking Square Roots: Ccss.Math.Content.Hsa Rei.B.4b

Solve $- 11 x^{2} + 20 x - 44 = 0$

Possible Answers:

$x = - \frac{12}{11} - \frac{8 i}{11} \sqrt{6} , x = \frac{230}{11} + \frac{8 i}{11} \sqrt{6}$

$x = \frac{120}{11} - \frac{8 i}{11} \sqrt{6} , x = - \frac{100}{11} + \frac{8 i}{11} \sqrt{6}$

$x = \frac{10}{11} - \frac{8 i}{11} \sqrt{6} , x = \frac{10}{11} + \frac{8 i}{11} \sqrt{6}$

$x = \frac{20}{11} - \frac{16 i}{11} \sqrt{6} , x = - \frac{20}{11} - \frac{16 i}{11} \sqrt{6}$

$x = \frac{21}{11} - \frac{8 i}{11} \sqrt{6} , x = - \frac{1}{11} + \frac{8 i}{11} \sqrt{6}$

Correct answer:

$x = \frac{10}{11} - \frac{8 i}{11} \sqrt{6} , x = \frac{10}{11} + \frac{8 i}{11} \sqrt{6}$

Explanation:

We can solve this by using the quadratic formula.

The quadratic formula is

$\frac{ -b \pm\sqrt{ b^{2} - 4 a c}}{2a}$

$\uptext{a}$ , $\uptext{b}$ , and $\uptext{c}$ correspond to coefficients in the quadratic equation, which is

$a x^{2} + b x - 44 = 0$

In this case a = -11 , b = 20 , and c = -44 .

$\frac{ -20 \pm\sqrt{ 20 ^2- 4 \cdot -11 \cdot -44 }}{ 2 \cdot -11 }$

$\frac{ -20 \pm\sqrt{ 400 - 1936 }}{ -22 }$

$\frac{ -20 \pm\sqrt{ -1536 }}{ -22 }$

Since the number inside the square root is negative, we will have an imaginary answer.

We will put an $\uptext{i}$ , outside the square root sign, and then do normal operations.

$\frac{ -20 \pm i \sqrt{ 1536 }}{ -22 }$

Now we split this up into two equations.

$\frac{ -20 + i \sqrt{ 1536 }}{ -22 }$

$\frac{10}{11} + \frac{8 i}{11} \sqrt{6}$

$\frac{ -20 - i \sqrt{ 1536 }}{ -22 }$

$\frac{10}{11} - \frac{8 i}{11} \sqrt{6}$

So our solutions are $x = \frac{10}{11} - \frac{8 i}{11} \sqrt{6}$ and $x = \frac{10}{11} + \frac{8 i}{11} \sqrt{6}$

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Example Question #2 : Solve Quadratic Equations By Inspection, Quadratic Formula, Factoring, Completing The Square, And Taking Square Roots: Ccss.Math.Content.Hsa Rei.B.4b

Solve $18 x^{2} - 14 x + 41 = 0$

Possible Answers:

$x = \frac{7}{9} - \frac{\sqrt{689} i}{9} , x = - \frac{7}{9} - \frac{\sqrt{689} i}{9}$

$x = \frac{187}{18} - \frac{\sqrt{689} i}{18} , x = - \frac{173}{18} + \frac{\sqrt{689} i}{18}$

$x = \frac{7}{18} - \frac{\sqrt{689} i}{18} , x = \frac{7}{18} + \frac{\sqrt{689} i}{18}$

$x = - \frac{29}{18} - \frac{\sqrt{689} i}{18} , x = \frac{367}{18} + \frac{\sqrt{689} i}{18}$

$x = \frac{25}{18} - \frac{\sqrt{689} i}{18} , x = - \frac{11}{18} + \frac{\sqrt{689} i}{18}$

Correct answer:

$x = \frac{7}{18} - \frac{\sqrt{689} i}{18} , x = \frac{7}{18} + \frac{\sqrt{689} i}{18}$

Explanation:

We can solve this by using the quadratic formula.

The quadratic formula is

$\frac{ -b \pm\sqrt{ b^{2} - 4 a c }}{ 2 a }$

$\uptext{a}$ , $\uptext{b}$ , and $\uptext{c}$ correspond to coefficients in the quadratic equation, which is

$a x^{2} + b x + 41 = 0$

In this case a = 18 , b = -14 , and c = 41 .

$\frac{ 14 \pm\sqrt{ -14 ^2- 4 \cdot 18 \cdot 41 }}{ 2 \cdot 18 }$

$\frac{ 14 \pm\sqrt{ 196 - 2952 }}{ 36 }$

$\frac{ 14 \pm\sqrt{ -2756 }}{ 36 }$

Since the number inside the square root is negative, we will have an imaginary answer.

We will put an $\uptext{i}$ , outside the square root sign, and then do normal operations.

$\frac{ 14 \pm i \sqrt{ 2756 }}{ 36 }$

Now we split this up into two equations.

$\frac{ 14 + i \sqrt{ 2756 }}{ 36 }$

$\frac{7}{18} + \frac{\sqrt{689} i}{18}$

$\frac{ 14 - i \sqrt{ 2756 }}{ 36 }$

$\frac{7}{18} - \frac{\sqrt{689} i}{18}$

So our solutions are $x = \frac{7}{18} - \frac{\sqrt{689} i}{18}$ and $x = \frac{7}{18} + \frac{\sqrt{689} i}{18}$

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All Common Core: High School - Algebra Resources

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Common Core: High School - Algebra : High School: Algebra

Study concepts, example questions & explanations for Common Core: High School - Algebra

All Common Core: High School - Algebra Resources

Example Questions

Example Question #501 : High School: Algebra

Example Question #502 : High School: Algebra

Example Question #503 : High School: Algebra

Example Question #504 : High School: Algebra

Example Question #505 : High School: Algebra

Example Question #506 : High School: Algebra

Example Question #507 : High School: Algebra

Example Question #501 : High School: Algebra

$x^{2} + 19 x - 52 = 0$

Example Question #1 : Solve Quadratic Equations By Inspection, Quadratic Formula, Factoring, Completing The Square, And Taking Square Roots: Ccss.Math.Content.Hsa Rei.B.4b

Example Question #2 : Solve Quadratic Equations By Inspection, Quadratic Formula, Factoring, Completing The Square, And Taking Square Roots: Ccss.Math.Content.Hsa Rei.B.4b

All Common Core: High School - Algebra Resources

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