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please help me find out the answer to this! 

 

4x2-31x=8

aolve for x.

You should get two answers. 

 Feb 16, 2018

Best Answer 

 #1
avatar+7354 
+4

\(4x^2 - 31x = 8\)

                                      Subtract  8  from both sides of the equation.

\(4x^2 - 31x - 8 = 0\)

 

Now we can solve this quadratic equation using the quadratic formula.

 

The quadratic formula says...

 

If     \(ax^2+bx+c=0\)    then    \(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

 

So...

 

If    \(4x^2 - 31x - 8 = 0\)    then    \(x = {-(-31) \pm \sqrt{(-31)^2-4(4)(-8)} \over 2(4)}\)

 

 

\(x = {-(-31) \pm \sqrt{(-31)^2-4(4)(-8)} \over 2(4)} \\~\\ x = {31 \pm \sqrt{961+128} \over 8} \\~\\ x = {31 \pm \sqrt{1089} \over 8} \\~\\ x = {31 \pm 33 \over 8} \\~\\ \begin{array} \ x=\frac{31+33}{8}\qquad&\text{or}&\qquad x=\frac{31-33}{8} \\~\\ x=\frac{64}{8}&\text{or}&\qquad x=\frac{-2}{8} \\~\\ x=8&\text{or}&\qquad x=-\frac{1}{4} \end{array}\)

.
 Feb 16, 2018
edited by hectictar  Feb 16, 2018
 #1
avatar+7354 
+4
Best Answer

\(4x^2 - 31x = 8\)

                                      Subtract  8  from both sides of the equation.

\(4x^2 - 31x - 8 = 0\)

 

Now we can solve this quadratic equation using the quadratic formula.

 

The quadratic formula says...

 

If     \(ax^2+bx+c=0\)    then    \(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

 

So...

 

If    \(4x^2 - 31x - 8 = 0\)    then    \(x = {-(-31) \pm \sqrt{(-31)^2-4(4)(-8)} \over 2(4)}\)

 

 

\(x = {-(-31) \pm \sqrt{(-31)^2-4(4)(-8)} \over 2(4)} \\~\\ x = {31 \pm \sqrt{961+128} \over 8} \\~\\ x = {31 \pm \sqrt{1089} \over 8} \\~\\ x = {31 \pm 33 \over 8} \\~\\ \begin{array} \ x=\frac{31+33}{8}\qquad&\text{or}&\qquad x=\frac{31-33}{8} \\~\\ x=\frac{64}{8}&\text{or}&\qquad x=\frac{-2}{8} \\~\\ x=8&\text{or}&\qquad x=-\frac{1}{4} \end{array}\)

hectictar Feb 16, 2018
edited by hectictar  Feb 16, 2018
 #2
avatar+23 
+2

Thank you so much. I am so grateful!!😊

FieryJade122  Feb 16, 2018

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