what are the solutions of the quadratic equation below? 3x2 - x = 11
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3x^2-x-11=0
disctirminant = (-1)^2-4.3. (-11)
disctirminant =1+132=133
x1= -(-1)+ square root of 133/2.3
x1= 1+ square root of 133/6
x2= -(-1) - square root of 133/2.3
x2= 1- square root of 133/6
Answer:
A quadratic equation \(ax^2+bx+c=0\) ....[1]
then the solution is given by:
\(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\)
Given the equation:
\(3x^2-x =11\)
Subtract 11 from both sides we have;
\(3x^2-x-11 =0\)
On comparing with [1] we have;
a = 3 , b =-1 and c =-11
Substitute these we have;
\(x = \frac{-(-1) \pm \sqrt{(-1)^2-4(3)(-11)}}{2(3)}\)
⇒\(x = \frac{1 \pm \sqrt{1+132}}{6}\)
⇒\(x = \frac{1 \pm \sqrt{133}}{6}\)
Therefore, the solution for the given equations are:
\(x = \frac{1 + \sqrt{133}}{6}\), \(\frac{1 -\sqrt{133}}{6}\)