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

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8 Diagnostic Tests 97 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #3 : Solve Quadratic Equations By Inspection, Quadratic Formula, Factoring, Completing The Square, And Taking Square Roots: Ccss.Math.Content.Hsa Rei.B.4b

Solve $22 x^{2} - 9 x + 16 = 0$

Possible Answers:

$x = \frac{53}{44} - \frac{\sqrt{1327} i}{44} , x = - \frac{35}{44} + \frac{\sqrt{1327} i}{44}$

$x = - \frac{79}{44} - \frac{\sqrt{1327} i}{44} , x = \frac{889}{44} + \frac{\sqrt{1327} i}{44}$

$x = \frac{9}{22} - \frac{\sqrt{1327} i}{22} , x = - \frac{9}{22} - \frac{\sqrt{1327} i}{22}$

$x = \frac{449}{44} - \frac{\sqrt{1327} i}{44} , x = - \frac{431}{44} + \frac{\sqrt{1327} i}{44}$

$x = \frac{9}{44} - \frac{\sqrt{1327} i}{44} , x = \frac{9}{44} + \frac{\sqrt{1327} i}{44}$

Correct answer:

$x = \frac{9}{44} - \frac{\sqrt{1327} i}{44} , x = \frac{9}{44} + \frac{\sqrt{1327} i}{44}$

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

$a x^{2} + b x + 16 = 0$

In this case a = 22 , b = -9 , and c = 16 .

$\frac{ 9 \pm\sqrt{ -9 ^2- 4 \cdot 22 \cdot 16 }}{ 2 \cdot 22 }$

$\frac{ 9 \pm\sqrt{ 81 - 1408 }}{ 44 }$

$\frac{ 9 \pm\sqrt{ -1327 }}{ 44 }$

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.

$\frac{ 9 \pm i \sqrt{ 1327 }}{ 44 }$

Now we split this up into two equations.

$\frac{ 9 + i \sqrt{ 1327 }}{ 44 }$

$\frac{9}{44} + \frac{\sqrt{1327} i}{44}$

$\frac{ 9 - i \sqrt{ 1327 }}{ 44 }$

$\frac{9}{44} - \frac{\sqrt{1327} i}{44}$

So our solutions are $x = \frac{9}{44} - \frac{\sqrt{1327} i}{44}$ and $x = \frac{9}{44} + \frac{\sqrt{1327} i}{44}$

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

Solve $- 49 x^{2} + 17 x - 45 = 0$

Possible Answers:

$x = \frac{17}{98} - \frac{\sqrt{8531} i}{98} , x = \frac{17}{98} + \frac{\sqrt{8531} i}{98}$

$x = \frac{17}{49} - \frac{\sqrt{8531} i}{49} , x = - \frac{17}{49} - \frac{\sqrt{8531} i}{49}$

$x = \frac{997}{98} - \frac{\sqrt{8531} i}{98} , x = - \frac{963}{98} + \frac{\sqrt{8531} i}{98}$

$x = \frac{115}{98} - \frac{\sqrt{8531} i}{98} , x = - \frac{81}{98} + \frac{\sqrt{8531} i}{98}$

$x = - \frac{179}{98} - \frac{\sqrt{8531} i}{98} , x = \frac{1977}{98} + \frac{\sqrt{8531} i}{98}$

Correct answer:

$x = \frac{17}{98} - \frac{\sqrt{8531} i}{98} , x = \frac{17}{98} + \frac{\sqrt{8531} i}{98}$

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

$a x^{2} + b x - 45 = 0$

In this case a = -49 , b = 17 , and c = -45 .

$\frac{ -17 \pm\sqrt{ 17 ^2- 4 \cdot -49 \cdot -45 }}{ 2 \cdot -49 }$

$\frac{ -17 \pm\sqrt{ 289 - 8820 }}{ -98 }$

$\frac{ -17 \pm\sqrt{ -8531 }}{ -98 }$

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.

$\frac{ -17 \pm i \sqrt{ 8531 }}{ -98 }$

Now we split this up into two equations.

$\frac{ -17 + i \sqrt{ 8531 }}{ -98 }$

$\frac{17}{98} + \frac{\sqrt{8531} i}{98}$

$\frac{ -17 - i \sqrt{ 8531 }}{ -98 }$

$\frac{17}{98} - \frac{\sqrt{8531} i}{98}$

So our solutions are $x = \frac{17}{98} - \frac{\sqrt{8531} i}{98}$ and $x = \frac{17}{98} + \frac{\sqrt{8531} i}{98}$

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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 $28 x^{2} + 30 x + 32 = 0$

Possible Answers:

$x = - \frac{15}{28} - \frac{\sqrt{671} i}{28} , x = - \frac{15}{28} + \frac{\sqrt{671} i}{28}$

$x = - \frac{15}{14} - \frac{\sqrt{671} i}{14} , x = \frac{15}{14} - \frac{\sqrt{671} i}{14}$

$x = \frac{13}{28} - \frac{\sqrt{671} i}{28} , x = - \frac{43}{28} + \frac{\sqrt{671} i}{28}$

$x = \frac{265}{28} - \frac{\sqrt{671} i}{28} , x = - \frac{295}{28} + \frac{\sqrt{671} i}{28}$

$x = - \frac{71}{28} - \frac{\sqrt{671} i}{28} , x = \frac{545}{28} + \frac{\sqrt{671} i}{28}$

Correct answer:

$x = - \frac{15}{28} - \frac{\sqrt{671} i}{28} , x = - \frac{15}{28} + \frac{\sqrt{671} i}{28}$

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

$a x^{2} + b x + 32 = 0$

In this case a = 28 , b = 30 , and c = 32 .

$\frac{ -30 \pm\sqrt{ 30 ^2- 4 \cdot 28 \cdot 32 }}{ 2 \cdot 28 }$

$\frac{ -30 \pm\sqrt{ 900 - 3584 }}{ 56 }$

$\frac{ -30 \pm\sqrt{ -2684 }}{ 56 }$

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.

$\frac{ -30 \pm i \sqrt{ 2684 }}{ 56 }$

Now we split this up into two equations.

$\frac{ -30 + i \sqrt{ 2684 }}{ 56 }$

$- \frac{15}{28} + \frac{\sqrt{671} i}{28}$

$\frac{ -30 - i \sqrt{ 2684 }}{ 56 }$

$- \frac{15}{28} - \frac{\sqrt{671} i}{28}$

So our solutions are $x = - \frac{15}{28} - \frac{\sqrt{671} i}{28}$ and $x = - \frac{15}{28} + \frac{\sqrt{671} i}{28}$

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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 $- 49 x^{2} + 24 x - 40 = 0$

Possible Answers:

$x = \frac{502}{49} - \frac{2 i}{49} \sqrt{454} , x = - \frac{478}{49} + \frac{2 i}{49} \sqrt{454}$

$x = \frac{24}{49} - \frac{4 i}{49} \sqrt{454} , x = - \frac{24}{49} - \frac{4 i}{49} \sqrt{454}$

$x = \frac{12}{49} - \frac{2 i}{49} \sqrt{454} , x = \frac{12}{49} + \frac{2 i}{49} \sqrt{454}$

$x = \frac{61}{49} - \frac{2 i}{49} \sqrt{454} , x = - \frac{37}{49} + \frac{2 i}{49} \sqrt{454}$

$x = - \frac{86}{49} - \frac{2 i}{49} \sqrt{454} , x = \frac{992}{49} + \frac{2 i}{49} \sqrt{454}$

Correct answer:

$x = \frac{12}{49} - \frac{2 i}{49} \sqrt{454} , x = \frac{12}{49} + \frac{2 i}{49} \sqrt{454}$

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

$a x^{2} + b x - 40 = 0$

In this case a = -49 , b = 24 , and c = -40 .

$\frac{ -24 \pm\sqrt{ 24 ^2- 4 \cdot -49 \cdot -40 }}{ 2 \cdot -49 }$

$\frac{ -24 \pm\sqrt{ 576 - 7840 }}{ -98 }$

$\frac{ -24 \pm\sqrt{ -7264 }}{ -98 }$

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.

$\frac{ -24 \pm i \sqrt{ 7264 }}{ -98 }$

Now we split this up into two equations.

$\frac{ -24 + i \sqrt{ 7264 }}{ -98 }$

$\frac{12}{49} + \frac{2 i}{49} \sqrt{454}$

$\frac{ -24 - i \sqrt{ 7264 }}{ -98 }$

$\frac{12}{49} - \frac{2 i}{49} \sqrt{454}$

So our solutions are $x = \frac{12}{49} - \frac{2 i}{49} \sqrt{454}$ and $x = \frac{12}{49} + \frac{2 i}{49} \sqrt{454}$

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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 $- 41 x^{2} + 46 x - 20 = 0$

Possible Answers:

$x = \frac{64}{41} - \frac{\sqrt{291} i}{41} , x = - \frac{18}{41} + \frac{\sqrt{291} i}{41}$

$x = \frac{433}{41} - \frac{\sqrt{291} i}{41} , x = - \frac{387}{41} + \frac{\sqrt{291} i}{41}$

$x = \frac{46}{41} - \frac{2 i}{41} \sqrt{291} , x = - \frac{46}{41} - \frac{2 i}{41} \sqrt{291}$

$x = \frac{23}{41} - \frac{\sqrt{291} i}{41} , x = \frac{23}{41} + \frac{\sqrt{291} i}{41}$

$x = - \frac{59}{41} - \frac{\sqrt{291} i}{41} , x = \frac{843}{41} + \frac{\sqrt{291} i}{41}$

Correct answer:

$x = \frac{23}{41} - \frac{\sqrt{291} i}{41} , x = \frac{23}{41} + \frac{\sqrt{291} i}{41}$

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

$a x^{2} + b x - 20 = 0$

In this case a = -41 , b = 46 , and c = -20 .

$\frac{ -46 \pm\sqrt{ 46 ^2- 4 \cdot -41 \cdot -20 }}{ 2 \cdot -41 }$

$\frac{ -46 \pm\sqrt{ 2116 - 3280 }}{ -82 }$

$\frac{ -46 \pm\sqrt{ -1164 }}{ -82 }$

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.

$\frac{ -46 \pm i \sqrt{ 1164 }}{ -82 }$

Now we split this up into two equations.

$\frac{ -46 + i \sqrt{ 1164 }}{ -82 }$

$\frac{23}{41} + \frac{\sqrt{291} i}{41}$

$\frac{ -46 - i \sqrt{ 1164 }}{ -82 }$

$\frac{23}{41} - \frac{\sqrt{291} i}{41}$

So our solutions are $x = \frac{23}{41} - \frac{\sqrt{291} i}{41}$ and $x = \frac{23}{41} + \frac{\sqrt{291} i}{41}$

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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 $- 12 x^{2} - 27 x - 30 = 0$

Possible Answers:

$x = - \frac{25}{8} - \frac{\sqrt{79} i}{8} , x = \frac{151}{8} + \frac{\sqrt{79} i}{8}$

$x = - \frac{9}{4} - \frac{\sqrt{79} i}{4} , x = \frac{9}{4} - \frac{\sqrt{79} i}{4}$

$x = - \frac{9}{8} - \frac{\sqrt{79} i}{8} , x = - \frac{9}{8} + \frac{\sqrt{79} i}{8}$

$x = \frac{71}{8} - \frac{\sqrt{79} i}{8} , x = - \frac{89}{8} + \frac{\sqrt{79} i}{8}$

$x = - \frac{1}{8} - \frac{\sqrt{79} i}{8} , x = - \frac{17}{8} + \frac{\sqrt{79} i}{8}$

Correct answer:

$x = - \frac{9}{8} - \frac{\sqrt{79} i}{8} , x = - \frac{9}{8} + \frac{\sqrt{79} i}{8}$

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

$a x^{2} + b x - 30 = 0$

In this case a = -12 , b = -27 , and c = -30 .

$\frac{ 27 \pm\sqrt{ -27 ^2- 4 \cdot -12 \cdot -30 }}{ 2 \cdot -12 }$

$\frac{ 27 \pm\sqrt{ 729 - 1440 }}{ -24 }$

$\frac{ 27 \pm\sqrt{ -711 }}{ -24 }$

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.

$\frac{ 27 \pm i \sqrt{ 711 }}{ -24 }$

Now we split this up into two equations.

$\frac{ 27 + i \sqrt{ 711 }}{ -24 }$

$- \frac{9}{8} + \frac{\sqrt{79} i}{8}$

$\frac{ 27 - i \sqrt{ 711 }}{ -24 }$

$- \frac{9}{8} - \frac{\sqrt{79} i}{8}$

So our solutions are $x = - \frac{9}{8} - \frac{\sqrt{79} i}{8}$ and $x = - \frac{9}{8} + \frac{\sqrt{79} i}{8}$

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Example Question #6 : Solve Quadratic Equations By Inspection, Quadratic Formula, Factoring, Completing The Square, And Taking Square Roots: Ccss.Math.Content.Hsa Rei.B.4b

Solve $- 23 x^{2} + 34 x - 44 = 0$

Possible Answers:

$x = \frac{40}{23} - \frac{\sqrt{723} i}{23} , x = - \frac{6}{23} + \frac{\sqrt{723} i}{23}$

$x = \frac{17}{23} - \frac{\sqrt{723} i}{23} , x = \frac{17}{23} + \frac{\sqrt{723} i}{23}$

$x = - \frac{29}{23} - \frac{\sqrt{723} i}{23} , x = \frac{477}{23} + \frac{\sqrt{723} i}{23}$

$x = \frac{34}{23} - \frac{2 i}{23} \sqrt{723} , x = - \frac{34}{23} - \frac{2 i}{23} \sqrt{723}$

$x = \frac{247}{23} - \frac{\sqrt{723} i}{23} , x = - \frac{213}{23} + \frac{\sqrt{723} i}{23}$

Correct answer:

$x = \frac{17}{23} - \frac{\sqrt{723} i}{23} , x = \frac{17}{23} + \frac{\sqrt{723} i}{23}$

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

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

In this case a = -23 , b = 34 , and c = -44 .

$\frac{ -34 \pm\sqrt{ 34 ^2- 4 \cdot -23 \cdot -44 }}{ 2 \cdot -23 }$

$\frac{ -34 \pm\sqrt{ 1156 - 4048 }}{ -46 }$

$\frac{ -34 \pm\sqrt{ -2892 }}{ -46 }$

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.

$\frac{ -34 \pm i \sqrt{ 2892 }}{ -46 }$

Now we split this up into two equations.

$\frac{ -34 + i \sqrt{ 2892 }}{ -46 }$

$\frac{17}{23} + \frac{\sqrt{723} i}{23}$

$\frac{ -34 - i \sqrt{ 2892 }}{ -46 }$

$\frac{17}{23} - \frac{\sqrt{723} i}{23}$

So our solutions are $x = \frac{17}{23} - \frac{\sqrt{723} i}{23}$ and $x = \frac{17}{23} + \frac{\sqrt{723} i}{23}$

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Example Question #7 : Solve Quadratic Equations By Inspection, Quadratic Formula, Factoring, Completing The Square, And Taking Square Roots: Ccss.Math.Content.Hsa Rei.B.4b

Solve $- 24 x^{2} + 17 x - 32 = 0$

Possible Answers:

$x = \frac{497}{48} - \frac{11 i}{48} \sqrt{23} , x = - \frac{463}{48} + \frac{11 i}{48} \sqrt{23}$

$x = - \frac{79}{48} - \frac{11 i}{48} \sqrt{23} , x = \frac{977}{48} + \frac{11 i}{48} \sqrt{23}$

$x = \frac{17}{24} - \frac{11 i}{24} \sqrt{23} , x = - \frac{17}{24} - \frac{11 i}{24} \sqrt{23}$

$x = \frac{17}{48} - \frac{11 i}{48} \sqrt{23} , x = \frac{17}{48} + \frac{11 i}{48} \sqrt{23}$

$x = \frac{65}{48} - \frac{11 i}{48} \sqrt{23} , x = - \frac{31}{48} + \frac{11 i}{48} \sqrt{23}$

Correct answer:

$x = \frac{17}{48} - \frac{11 i}{48} \sqrt{23} , x = \frac{17}{48} + \frac{11 i}{48} \sqrt{23}$

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

$a x^{2} + b x - 32 = 0$

In this case a = -24 , b = 17 , and c = -32 .

$\frac{ -17 \pm\sqrt{ 17 ^2- 4 \cdot -24 \cdot -32 }}{ 2 \cdot -24 }$

$\frac{ -17 \pm\sqrt{ 289 - 3072 }}{ -48 }$

$\frac{ -17 \pm\sqrt{ -2783 }}{ -48 }$

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.

$\frac{ -17 \pm i \sqrt{ 2783 }}{ -48 }$

Now we split this up into two equations.

$\frac{ -17 + i \sqrt{ 2783 }}{ -48 }$

$\frac{17}{48} + \frac{11 i}{48} \sqrt{23}$

$\frac{ -17 - i \sqrt{ 2783 }}{ -48 }$

$\frac{17}{48} - \frac{11 i}{48} \sqrt{23}$

So our solutions are $x = \frac{17}{48} - \frac{11 i}{48} \sqrt{23}$ and $x = \frac{17}{48} + \frac{11 i}{48} \sqrt{23}$

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Example Question #11 : Solve Quadratic Equations By Inspection, Quadratic Formula, Factoring, Completing The Square, And Taking Square Roots: Ccss.Math.Content.Hsa Rei.B.4b

Solve $27 x^{2} + 3 x + 5 = 0$

Possible Answers:

$x = - \frac{37}{18} - \frac{\sqrt{59} i}{18} , x = \frac{359}{18} + \frac{\sqrt{59} i}{18}$

$x = - \frac{1}{9} - \frac{\sqrt{59} i}{9} , x = \frac{1}{9} - \frac{\sqrt{59} i}{9}$

$x = - \frac{1}{18} - \frac{\sqrt{59} i}{18} , x = - \frac{1}{18} + \frac{\sqrt{59} i}{18}$

$x = \frac{179}{18} - \frac{\sqrt{59} i}{18} , x = - \frac{181}{18} + \frac{\sqrt{59} i}{18}$

$x = \frac{17}{18} - \frac{\sqrt{59} i}{18} , x = - \frac{19}{18} + \frac{\sqrt{59} i}{18}$

Correct answer:

$x = - \frac{1}{18} - \frac{\sqrt{59} i}{18} , x = - \frac{1}{18} + \frac{\sqrt{59} i}{18}$

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

$a x^{2} + b x + 5 = 0$

In this case a = 27 , b = 3 , and c = 5 .

$\frac{ -3 \pm\sqrt{ 3 ^2- 4 \cdot 27 \cdot 5 }}{ 2 \cdot 27 }$

$\frac{ -3 \pm\sqrt{ 9 - 540 }}{ 54 }$

$\frac{ -3 \pm\sqrt{ -531 }}{ 54 }$

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.

$\frac{ -3 \pm i \sqrt{ 531 }}{ 54 }$

Now we split this up into two equations.

$\frac{ -3 + i \sqrt{ 531 }}{ 54 }$

$- \frac{1}{18} + \frac{\sqrt{59} i}{18}$

$\frac{ -3 - i \sqrt{ 531 }}{ 54 }$

$- \frac{1}{18} - \frac{\sqrt{59} i}{18}$

So our solutions are $x = - \frac{1}{18} - \frac{\sqrt{59} i}{18}$ and $x = - \frac{1}{18} + \frac{\sqrt{59} i}{18}$

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Example Question #61 : Reasoning With Equations & Inequalities

Solve $12 x^{2} - 13 x + 41 = 0$

Possible Answers:

$x = \frac{37}{24} - \frac{\sqrt{1799} i}{24} , x = - \frac{11}{24} + \frac{\sqrt{1799} i}{24}$

$x = - \frac{35}{24} - \frac{\sqrt{1799} i}{24} , x = \frac{493}{24} + \frac{\sqrt{1799} i}{24}$

$x = \frac{13}{24} - \frac{\sqrt{1799} i}{24} , x = \frac{13}{24} + \frac{\sqrt{1799} i}{24}$

$x = \frac{253}{24} - \frac{\sqrt{1799} i}{24} , x = - \frac{227}{24} + \frac{\sqrt{1799} i}{24}$

$x = \frac{13}{12} - \frac{\sqrt{1799} i}{12} , x = - \frac{13}{12} - \frac{\sqrt{1799} i}{12}$

Correct answer:

$x = \frac{13}{24} - \frac{\sqrt{1799} i}{24} , x = \frac{13}{24} + \frac{\sqrt{1799} i}{24}$

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 = 12 , b = -13 , and c = 41 .

$\frac{ 13 \pm\sqrt{ -13 ^2- 4 \cdot 12 \cdot 41 }}{ 2 \cdot 12 }$

$\frac{ 13 \pm\sqrt{ 169 - 1968 }}{ 24 }$

$\frac{ 13 \pm\sqrt{ -1799 }}{ 24 }$

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.

$\frac{ 13 \pm i \sqrt{ 1799 }}{ 24 }$

Now we split this up into two equations.

$\frac{ 13 + i \sqrt{ 1799 }}{ 24 }$

$\frac{13}{24} + \frac{\sqrt{1799} i}{24}$

$\frac{ 13 - i \sqrt{ 1799 }}{ 24 }$

$\frac{13}{24} - \frac{\sqrt{1799} i}{24}$

So our solutions are $x = \frac{13}{24} - \frac{\sqrt{1799} i}{24}$ and $x = \frac{13}{24} + \frac{\sqrt{1799} i}{24}$

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

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

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 #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 #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

Example Question #6 : Solve Quadratic Equations By Inspection, Quadratic Formula, Factoring, Completing The Square, And Taking Square Roots: Ccss.Math.Content.Hsa Rei.B.4b

Example Question #7 : Solve Quadratic Equations By Inspection, Quadratic Formula, Factoring, Completing The Square, And Taking Square Roots: Ccss.Math.Content.Hsa Rei.B.4b

Example Question #11 : Solve Quadratic Equations By Inspection, Quadratic Formula, Factoring, Completing The Square, And Taking Square Roots: Ccss.Math.Content.Hsa Rei.B.4b

Example Question #61 : Reasoning With Equations & Inequalities

All Common Core: High School - Algebra Resources

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