+31 please help me find out the answer to this!
4x2-31x=8
aolve for x.
You should get two answers.
+9481 \(4x^2 - 31x = 8\)
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 \(x = {-(-31) \pm \sqrt{(-31)^2-4(4)(-8)} \over 2(4)}\)
\(x = {-(-31) \pm \sqrt{(-31)^2-4(4)(-8)} \over 2(4)} \\~\\ x = {31 \pm \sqrt{961+128} \over 8} \\~\\ x = {31 \pm \sqrt{1089} \over 8} \\~\\ x = {31 \pm 33 \over 8} \\~\\ \begin{array} \ x=\frac{31+33}{8}\qquad&\text{or}&\qquad x=\frac{31-33}{8} \\~\\ x=\frac{64}{8}&\text{or}&\qquad x=\frac{-2}{8} \\~\\ x=8&\text{or}&\qquad x=-\frac{1}{4} \end{array}\)
+9481 \(4x^2 - 31x = 8\)
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 \(x = {-(-31) \pm \sqrt{(-31)^2-4(4)(-8)} \over 2(4)}\)
\(x = {-(-31) \pm \sqrt{(-31)^2-4(4)(-8)} \over 2(4)} \\~\\ x = {31 \pm \sqrt{961+128} \over 8} \\~\\ x = {31 \pm \sqrt{1089} \over 8} \\~\\ x = {31 \pm 33 \over 8} \\~\\ \begin{array} \ x=\frac{31+33}{8}\qquad&\text{or}&\qquad x=\frac{31-33}{8} \\~\\ x=\frac{64}{8}&\text{or}&\qquad x=\frac{-2}{8} \\~\\ x=8&\text{or}&\qquad x=-\frac{1}{4} \end{array}\)
