Common Core: High School - Algebra : Solve Quadratic Equations by Inspection, Quadratic Formula, Factoring, Completing the Square, and Taking Square Roots: CCSS.Math.Content.HSA-REI.B.4b

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

varsity tutors app store varsity tutors android store

All Common Core: High School - Algebra Resources

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

Example Questions

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 \(\displaystyle - 11 x^{2} + 20 x - 44 = 0\)

Possible Answers:

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

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

\(\displaystyle x = \frac{10}{11} - \frac{8 i}{11} \sqrt{6} , x = \frac{10}{11} + \frac{8 i}{11} \sqrt{6}\)

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

\(\displaystyle x = \frac{20}{11} - \frac{16 i}{11} \sqrt{6} , x = - \frac{20}{11} - \frac{16 i}{11} \sqrt{6}\)

Correct answer:

\(\displaystyle x = \frac{10}{11} - \frac{8 i}{11} \sqrt{6} , x = \frac{10}{11} + \frac{8 i}{11} \sqrt{6}\)

Explanation:

We can solve this by using the quadratic formula.

The quadratic formula is

\(\displaystyle \frac{ -b \pm\sqrt{ b^{2} - 4 a c}}{2a}\)

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

\(\displaystyle a x^{2} + b x - 44 = 0\)

In this case \(\displaystyle a = -11\)\(\displaystyle b = 20\), and \(\displaystyle c = -44\) .

\(\displaystyle \frac{ -20 \pm\sqrt{ 20 ^2- 4 \cdot -11 \cdot -44 }}{ 2 \cdot -11 }\)

\(\displaystyle \frac{ -20 \pm\sqrt{ 400 - 1936 }}{ -22 }\)

\(\displaystyle \frac{ -20 \pm\sqrt{ -1536 }}{ -22 }\)

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

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

\(\displaystyle \frac{ -20 \pm i \sqrt{ 1536 }}{ -22 }\)

Now we split this up into two equations.

\(\displaystyle \frac{ -20 + i \sqrt{ 1536 }}{ -22 }\)

\(\displaystyle \frac{10}{11} + \frac{8 i}{11} \sqrt{6}\)

\(\displaystyle \frac{ -20 - i \sqrt{ 1536 }}{ -22 }\)

\(\displaystyle \frac{10}{11} - \frac{8 i}{11} \sqrt{6}\)

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

 

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 \(\displaystyle 18 x^{2} - 14 x + 41 = 0\)

Possible Answers:

\(\displaystyle x = \frac{25}{18} - \frac{\sqrt{689} i}{18} , x = - \frac{11}{18} + \frac{\sqrt{689} i}{18}\)

\(\displaystyle x = \frac{7}{18} - \frac{\sqrt{689} i}{18} , x = \frac{7}{18} + \frac{\sqrt{689} i}{18}\)

\(\displaystyle x = \frac{187}{18} - \frac{\sqrt{689} i}{18} , x = - \frac{173}{18} + \frac{\sqrt{689} i}{18}\)

\(\displaystyle x = \frac{7}{9} - \frac{\sqrt{689} i}{9} , x = - \frac{7}{9} - \frac{\sqrt{689} i}{9}\)

\(\displaystyle x = - \frac{29}{18} - \frac{\sqrt{689} i}{18} , x = \frac{367}{18} + \frac{\sqrt{689} i}{18}\)

Correct answer:

\(\displaystyle x = \frac{7}{18} - \frac{\sqrt{689} i}{18} , x = \frac{7}{18} + \frac{\sqrt{689} i}{18}\)

Explanation:

We can solve this by using the quadratic formula.

The quadratic formula is

\(\displaystyle \frac{ -b \pm\sqrt{ b^{2} - 4 a c }}{ 2 a }\)

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

\(\displaystyle a x^{2} + b x + 41 = 0\)

In this case \(\displaystyle a = 18\) , \(\displaystyle b = -14\) , and \(\displaystyle c = 41\) .

\(\displaystyle \frac{ 14 \pm\sqrt{ -14 ^2- 4 \cdot 18 \cdot 41 }}{ 2 \cdot 18 }\)

\(\displaystyle \frac{ 14 \pm\sqrt{ 196 - 2952 }}{ 36 }\)

\(\displaystyle \frac{ 14 \pm\sqrt{ -2756 }}{ 36 }\)

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

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

\(\displaystyle \frac{ 14 \pm i \sqrt{ 2756 }}{ 36 }\)

Now we split this up into two equations.

\(\displaystyle \frac{ 14 + i \sqrt{ 2756 }}{ 36 }\)

\(\displaystyle \frac{7}{18} + \frac{\sqrt{689} i}{18}\)

\(\displaystyle \frac{ 14 - i \sqrt{ 2756 }}{ 36 }\)

\(\displaystyle \frac{7}{18} - \frac{\sqrt{689} i}{18}\)

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

 

Example Question #511 : High School: Algebra

Solve \(\displaystyle 22 x^{2} - 9 x + 16 = 0\)

Possible Answers:

\(\displaystyle x = \frac{449}{44} - \frac{\sqrt{1327} i}{44} , x = - \frac{431}{44} + \frac{\sqrt{1327} i}{44}\)

\(\displaystyle x = - \frac{79}{44} - \frac{\sqrt{1327} i}{44} , x = \frac{889}{44} + \frac{\sqrt{1327} i}{44}\)

\(\displaystyle x = \frac{9}{44} - \frac{\sqrt{1327} i}{44} , x = \frac{9}{44} + \frac{\sqrt{1327} i}{44}\)

\(\displaystyle x = \frac{53}{44} - \frac{\sqrt{1327} i}{44} , x = - \frac{35}{44} + \frac{\sqrt{1327} i}{44}\)

\(\displaystyle x = \frac{9}{22} - \frac{\sqrt{1327} i}{22} , x = - \frac{9}{22} - \frac{\sqrt{1327} i}{22}\)

Correct answer:

\(\displaystyle x = \frac{9}{44} - \frac{\sqrt{1327} i}{44} , x = \frac{9}{44} + \frac{\sqrt{1327} i}{44}\)

Explanation:

We can solve this by using the quadratic formula.

The quadratic formula is

\(\displaystyle \frac{ -b \pm\sqrt{ b^{2} - 4 a c }}{ 2 a }\)

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

\(\displaystyle a x^{2} + b x + 16 = 0\)

In this case \(\displaystyle a = 22\) , \(\displaystyle b = -9\) , and \(\displaystyle c = 16\) .

\(\displaystyle \frac{ 9 \pm\sqrt{ -9 ^2- 4 \cdot 22 \cdot 16 }}{ 2 \cdot 22 }\)

\(\displaystyle \frac{ 9 \pm\sqrt{ 81 - 1408 }}{ 44 }\)

\(\displaystyle \frac{ 9 \pm\sqrt{ -1327 }}{ 44 }\)

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

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

\(\displaystyle \frac{ 9 \pm i \sqrt{ 1327 }}{ 44 }\)

Now we split this up into two equations.

\(\displaystyle \frac{ 9 + i \sqrt{ 1327 }}{ 44 }\)

\(\displaystyle \frac{9}{44} + \frac{\sqrt{1327} i}{44}\)

\(\displaystyle \frac{ 9 - i \sqrt{ 1327 }}{ 44 }\)

\(\displaystyle \frac{9}{44} - \frac{\sqrt{1327} i}{44}\)

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

 

Example Question #512 : High School: Algebra

Solve \(\displaystyle - 49 x^{2} + 17 x - 45 = 0\)

Possible Answers:

\(\displaystyle x = \frac{997}{98} - \frac{\sqrt{8531} i}{98} , x = - \frac{963}{98} + \frac{\sqrt{8531} i}{98}\)

\(\displaystyle x = \frac{115}{98} - \frac{\sqrt{8531} i}{98} , x = - \frac{81}{98} + \frac{\sqrt{8531} i}{98}\)

\(\displaystyle x = \frac{17}{49} - \frac{\sqrt{8531} i}{49} , x = - \frac{17}{49} - \frac{\sqrt{8531} i}{49}\)

\(\displaystyle x = - \frac{179}{98} - \frac{\sqrt{8531} i}{98} , x = \frac{1977}{98} + \frac{\sqrt{8531} i}{98}\)

\(\displaystyle x = \frac{17}{98} - \frac{\sqrt{8531} i}{98} , x = \frac{17}{98} + \frac{\sqrt{8531} i}{98}\)

Correct answer:

\(\displaystyle x = \frac{17}{98} - \frac{\sqrt{8531} i}{98} , x = \frac{17}{98} + \frac{\sqrt{8531} i}{98}\)

Explanation:

We can solve this by using the quadratic formula.

The quadratic formula is

\(\displaystyle \frac{ -b \pm\sqrt{ b^{2} - 4 a c }}{ 2 a }\)

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

\(\displaystyle a x^{2} + b x - 45 = 0\)

In this case \(\displaystyle a = -49\) , \(\displaystyle b = 17\) , and \(\displaystyle c = -45\).

\(\displaystyle \frac{ -17 \pm\sqrt{ 17 ^2- 4 \cdot -49 \cdot -45 }}{ 2 \cdot -49 }\)

\(\displaystyle \frac{ -17 \pm\sqrt{ 289 - 8820 }}{ -98 }\)

\(\displaystyle \frac{ -17 \pm\sqrt{ -8531 }}{ -98 }\)

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

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

\(\displaystyle \frac{ -17 \pm i \sqrt{ 8531 }}{ -98 }\)

Now we split this up into two equations.

\(\displaystyle \frac{ -17 + i \sqrt{ 8531 }}{ -98 }\)

\(\displaystyle \frac{17}{98} + \frac{\sqrt{8531} i}{98}\)

\(\displaystyle \frac{ -17 - i \sqrt{ 8531 }}{ -98 }\)

\(\displaystyle \frac{17}{98} - \frac{\sqrt{8531} i}{98}\)

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

 

Example Question #513 : High School: Algebra

Solve \(\displaystyle 28 x^{2} + 30 x + 32 = 0\)

Possible Answers:

\(\displaystyle x = - \frac{71}{28} - \frac{\sqrt{671} i}{28} , x = \frac{545}{28} + \frac{\sqrt{671} i}{28}\)

\(\displaystyle x = \frac{13}{28} - \frac{\sqrt{671} i}{28} , x = - \frac{43}{28} + \frac{\sqrt{671} i}{28}\)

\(\displaystyle x = - \frac{15}{28} - \frac{\sqrt{671} i}{28} , x = - \frac{15}{28} + \frac{\sqrt{671} i}{28}\)

\(\displaystyle x = - \frac{15}{14} - \frac{\sqrt{671} i}{14} , x = \frac{15}{14} - \frac{\sqrt{671} i}{14}\)

\(\displaystyle x = \frac{265}{28} - \frac{\sqrt{671} i}{28} , x = - \frac{295}{28} + \frac{\sqrt{671} i}{28}\)

Correct answer:

\(\displaystyle x = - \frac{15}{28} - \frac{\sqrt{671} i}{28} , x = - \frac{15}{28} + \frac{\sqrt{671} i}{28}\)

Explanation:

We can solve this by using the quadratic formula.

The quadratic formula is

\(\displaystyle \frac{ -b \pm\sqrt{ b^{2} - 4 a c }}{ 2 a }\)

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

\(\displaystyle a x^{2} + b x + 32 = 0\)

In this case \(\displaystyle a = 28\) , \(\displaystyle b = 30\) , and \(\displaystyle c = 32\).

\(\displaystyle \frac{ -30 \pm\sqrt{ 30 ^2- 4 \cdot 28 \cdot 32 }}{ 2 \cdot 28 }\)

\(\displaystyle \frac{ -30 \pm\sqrt{ 900 - 3584 }}{ 56 }\)

\(\displaystyle \frac{ -30 \pm\sqrt{ -2684 }}{ 56 }\)

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

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

\(\displaystyle \frac{ -30 \pm i \sqrt{ 2684 }}{ 56 }\)

Now we split this up into two equations.

\(\displaystyle \frac{ -30 + i \sqrt{ 2684 }}{ 56 }\)

\(\displaystyle - \frac{15}{28} + \frac{\sqrt{671} i}{28}\)

\(\displaystyle \frac{ -30 - i \sqrt{ 2684 }}{ 56 }\)

\(\displaystyle - \frac{15}{28} - \frac{\sqrt{671} i}{28}\)

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

 

Example Question #514 : High School: Algebra

Solve \(\displaystyle - 49 x^{2} + 24 x - 40 = 0\)

Possible Answers:

\(\displaystyle x = \frac{24}{49} - \frac{4 i}{49} \sqrt{454} , x = - \frac{24}{49} - \frac{4 i}{49} \sqrt{454}\)

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

\(\displaystyle x = \frac{12}{49} - \frac{2 i}{49} \sqrt{454} , x = \frac{12}{49} + \frac{2 i}{49} \sqrt{454}\)

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

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

Correct answer:

\(\displaystyle x = \frac{12}{49} - \frac{2 i}{49} \sqrt{454} , x = \frac{12}{49} + \frac{2 i}{49} \sqrt{454}\)

Explanation:

We can solve this by using the quadratic formula.

The quadratic formula is

\(\displaystyle \frac{ -b \pm\sqrt{ b^{2} - 4 a c }}{ 2 a }\)

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

\(\displaystyle a x^{2} + b x - 40 = 0\)

In this case \(\displaystyle a = -49\) , \(\displaystyle b = 24\), and \(\displaystyle c = -40\) .

\(\displaystyle \frac{ -24 \pm\sqrt{ 24 ^2- 4 \cdot -49 \cdot -40 }}{ 2 \cdot -49 }\)

\(\displaystyle \frac{ -24 \pm\sqrt{ 576 - 7840 }}{ -98 }\)

\(\displaystyle \frac{ -24 \pm\sqrt{ -7264 }}{ -98 }\)

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

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

\(\displaystyle \frac{ -24 \pm i \sqrt{ 7264 }}{ -98 }\)

Now we split this up into two equations.

\(\displaystyle \frac{ -24 + i \sqrt{ 7264 }}{ -98 }\)

\(\displaystyle \frac{12}{49} + \frac{2 i}{49} \sqrt{454}\)

\(\displaystyle \frac{ -24 - i \sqrt{ 7264 }}{ -98 }\)

\(\displaystyle \frac{12}{49} - \frac{2 i}{49} \sqrt{454}\)

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

 

Example Question #515 : High School: Algebra

Solve \(\displaystyle - 41 x^{2} + 46 x - 20 = 0\)

Possible Answers:

\(\displaystyle x = \frac{64}{41} - \frac{\sqrt{291} i}{41} , x = - \frac{18}{41} + \frac{\sqrt{291} i}{41}\)

\(\displaystyle x = \frac{46}{41} - \frac{2 i}{41} \sqrt{291} , x = - \frac{46}{41} - \frac{2 i}{41} \sqrt{291}\)

\(\displaystyle x = \frac{23}{41} - \frac{\sqrt{291} i}{41} , x = \frac{23}{41} + \frac{\sqrt{291} i}{41}\)

\(\displaystyle x = - \frac{59}{41} - \frac{\sqrt{291} i}{41} , x = \frac{843}{41} + \frac{\sqrt{291} i}{41}\)

\(\displaystyle x = \frac{433}{41} - \frac{\sqrt{291} i}{41} , x = - \frac{387}{41} + \frac{\sqrt{291} i}{41}\)

Correct answer:

\(\displaystyle x = \frac{23}{41} - \frac{\sqrt{291} i}{41} , x = \frac{23}{41} + \frac{\sqrt{291} i}{41}\)

Explanation:

We can solve this by using the quadratic formula.

The quadratic formula is

\(\displaystyle \frac{ -b \pm\sqrt{ b^{2} - 4 a c }}{ 2 a }\)

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

\(\displaystyle a x^{2} + b x - 20 = 0\)

In this case \(\displaystyle a = -41\) , \(\displaystyle b = 46\), and \(\displaystyle c = -20\).

\(\displaystyle \frac{ -46 \pm\sqrt{ 46 ^2- 4 \cdot -41 \cdot -20 }}{ 2 \cdot -41 }\)

\(\displaystyle \frac{ -46 \pm\sqrt{ 2116 - 3280 }}{ -82 }\)

\(\displaystyle \frac{ -46 \pm\sqrt{ -1164 }}{ -82 }\)

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.

\(\displaystyle \frac{ -46 \pm i \sqrt{ 1164 }}{ -82 }\)

Now we split this up into two equations.

\(\displaystyle \frac{ -46 + i \sqrt{ 1164 }}{ -82 }\)

\(\displaystyle \frac{23}{41} + \frac{\sqrt{291} i}{41}\)

\(\displaystyle \frac{ -46 - i \sqrt{ 1164 }}{ -82 }\)

\(\displaystyle \frac{23}{41} - \frac{\sqrt{291} i}{41}\)

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

 

Example Question #516 : High School: Algebra

Solve \(\displaystyle - 12 x^{2} - 27 x - 30 = 0\)

Possible Answers:

\(\displaystyle x = - \frac{9}{4} - \frac{\sqrt{79} i}{4} , x = \frac{9}{4} - \frac{\sqrt{79} i}{4}\)

\(\displaystyle x = - \frac{25}{8} - \frac{\sqrt{79} i}{8} , x = \frac{151}{8} + \frac{\sqrt{79} i}{8}\)

\(\displaystyle x = - \frac{1}{8} - \frac{\sqrt{79} i}{8} , x = - \frac{17}{8} + \frac{\sqrt{79} i}{8}\)

\(\displaystyle x = \frac{71}{8} - \frac{\sqrt{79} i}{8} , x = - \frac{89}{8} + \frac{\sqrt{79} i}{8}\)

\(\displaystyle x = - \frac{9}{8} - \frac{\sqrt{79} i}{8} , x = - \frac{9}{8} + \frac{\sqrt{79} i}{8}\)

Correct answer:

\(\displaystyle x = - \frac{9}{8} - \frac{\sqrt{79} i}{8} , x = - \frac{9}{8} + \frac{\sqrt{79} i}{8}\)

Explanation:

We can solve this by using the quadratic formula.

The quadratic formula is

\(\displaystyle \frac{ -b \pm\sqrt{ b^{2} - 4 a c }}{ 2 a }\)

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

\(\displaystyle a x^{2} + b x - 30 = 0\)

In this case \(\displaystyle a = -12\) , \(\displaystyle b = -27\) , and \(\displaystyle c = -30\) .

\(\displaystyle \frac{ 27 \pm\sqrt{ -27 ^2- 4 \cdot -12 \cdot -30 }}{ 2 \cdot -12 }\)

\(\displaystyle \frac{ 27 \pm\sqrt{ 729 - 1440 }}{ -24 }\)

\(\displaystyle \frac{ 27 \pm\sqrt{ -711 }}{ -24 }\)

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

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

\(\displaystyle \frac{ 27 \pm i \sqrt{ 711 }}{ -24 }\)

Now we split this up into two equations.

\(\displaystyle \frac{ 27 + i \sqrt{ 711 }}{ -24 }\)

\(\displaystyle - \frac{9}{8} + \frac{\sqrt{79} i}{8}\)

\(\displaystyle \frac{ 27 - i \sqrt{ 711 }}{ -24 }\)

\(\displaystyle - \frac{9}{8} - \frac{\sqrt{79} i}{8}\)

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

 

Example Question #517 : High School: Algebra

Solve \(\displaystyle - 23 x^{2} + 34 x - 44 = 0\)

Possible Answers:

\(\displaystyle x = \frac{34}{23} - \frac{2 i}{23} \sqrt{723} , x = - \frac{34}{23} - \frac{2 i}{23} \sqrt{723}\)

\(\displaystyle x = - \frac{29}{23} - \frac{\sqrt{723} i}{23} , x = \frac{477}{23} + \frac{\sqrt{723} i}{23}\)

\(\displaystyle x = \frac{17}{23} - \frac{\sqrt{723} i}{23} , x = \frac{17}{23} + \frac{\sqrt{723} i}{23}\)

\(\displaystyle x = \frac{247}{23} - \frac{\sqrt{723} i}{23} , x = - \frac{213}{23} + \frac{\sqrt{723} i}{23}\)

\(\displaystyle x = \frac{40}{23} - \frac{\sqrt{723} i}{23} , x = - \frac{6}{23} + \frac{\sqrt{723} i}{23}\)

Correct answer:

\(\displaystyle x = \frac{17}{23} - \frac{\sqrt{723} i}{23} , x = \frac{17}{23} + \frac{\sqrt{723} i}{23}\)

Explanation:

We can solve this by using the quadratic formula.

The quadratic formula is

\(\displaystyle \frac{ -b \pm\sqrt{ b^{2} - 4 a c }}{ 2 a }\)

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

\(\displaystyle a x^{2} + b x - 44 = 0\)

In this case \(\displaystyle a = -23\), \(\displaystyle b = 34\), and \(\displaystyle c = -44\).

\(\displaystyle \frac{ -34 \pm\sqrt{ 34 ^2- 4 \cdot -23 \cdot -44 }}{ 2 \cdot -23 }\)

\(\displaystyle \frac{ -34 \pm\sqrt{ 1156 - 4048 }}{ -46 }\)

\(\displaystyle \frac{ -34 \pm\sqrt{ -2892 }}{ -46 }\)

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

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

\(\displaystyle \frac{ -34 \pm i \sqrt{ 2892 }}{ -46 }\)

Now we split this up into two equations.

\(\displaystyle \frac{ -34 + i \sqrt{ 2892 }}{ -46 }\)

\(\displaystyle \frac{17}{23} + \frac{\sqrt{723} i}{23}\)

\(\displaystyle \frac{ -34 - i \sqrt{ 2892 }}{ -46 }\)

\(\displaystyle \frac{17}{23} - \frac{\sqrt{723} i}{23}\)

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

 

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 \(\displaystyle - 24 x^{2} + 17 x - 32 = 0\)

Possible Answers:

\(\displaystyle x = \frac{17}{24} - \frac{11 i}{24} \sqrt{23} , x = - \frac{17}{24} - \frac{11 i}{24} \sqrt{23}\)

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

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

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

\(\displaystyle x = \frac{17}{48} - \frac{11 i}{48} \sqrt{23} , x = \frac{17}{48} + \frac{11 i}{48} \sqrt{23}\)

Correct answer:

\(\displaystyle x = \frac{17}{48} - \frac{11 i}{48} \sqrt{23} , x = \frac{17}{48} + \frac{11 i}{48} \sqrt{23}\)

Explanation:

We can solve this by using the quadratic formula.

The quadratic formula is

\(\displaystyle \frac{ -b \pm\sqrt{ b^{2} - 4 a c }}{ 2 a }\)

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

\(\displaystyle a x^{2} + b x - 32 = 0\)

In this case \(\displaystyle a = -24\) , \(\displaystyle b = 17\) , and \(\displaystyle c = -32\).

\(\displaystyle \frac{ -17 \pm\sqrt{ 17 ^2- 4 \cdot -24 \cdot -32 }}{ 2 \cdot -24 }\)

\(\displaystyle \frac{ -17 \pm\sqrt{ 289 - 3072 }}{ -48 }\)

\(\displaystyle \frac{ -17 \pm\sqrt{ -2783 }}{ -48 }\)

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

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

\(\displaystyle \frac{ -17 \pm i \sqrt{ 2783 }}{ -48 }\)

Now we split this up into two equations.

\(\displaystyle \frac{ -17 + i \sqrt{ 2783 }}{ -48 }\)

\(\displaystyle \frac{17}{48} + \frac{11 i}{48} \sqrt{23}\)

\(\displaystyle \frac{ -17 - i \sqrt{ 2783 }}{ -48 }\)

\(\displaystyle \frac{17}{48} - \frac{11 i}{48} \sqrt{23}\)

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

 

All Common Core: High School - Algebra Resources

8 Diagnostic Tests 97 Practice Tests Question of the Day Flashcards Learn by Concept
Learning Tools by Varsity Tutors