Question

How do you solve $2{{x}^{2}}=x+8$?

Answer 1

Hint: Now to solve the equation we will first write the equation in general form $a{{x}^{2}}+bx+c$ Now we will divide the equation by a and then add and subtract ${{\left( \dfrac{b}{2a} \right)}^{2}}$ to the equation then we know that ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$ hence using this we will simplify the equation and shift the constants on RHS. Now we will take the square root of the equation and simplify to find the value of x.

Complete step by step solution:
Now we are given with a quadratic equation $2{{x}^{2}}=x+8$
Now first let us write the equation in general form which is $a{{x}^{2}}+bx+c$
Now by transposing the terms on RHS to LHS we get,
$\Rightarrow 2{{x}^{2}}-x-8=0$
Hence we get the equation in the form $a{{x}^{2}}+bx+c$ where a = 2, b = - 1 and c = - 8.
Now we will solve the equation by completing the square.
First we will make the coefficient of ${{x}^{2}}$ as 1 by dividing the whole equation by 2.
Hence we get,
$\Rightarrow {{x}^{2}}-\dfrac{1}{2}x-4=0$
Now we will add ${{\left( \dfrac{b}{2a} \right)}^{2}}$
Here b = - 1 and a = 2 hence ${{\left( \dfrac{b}{2a} \right)}^{2}}=\dfrac{1}{16}$
$\Rightarrow {{x}^{2}}-\dfrac{1}{2}x+\left( \dfrac{1}{16} \right)-\left( \dfrac{1}{16} \right)-4=0$
Now we know that ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$ . Hence we get,
$\Rightarrow {{\left( x-\dfrac{1}{4} \right)}^{2}}-\left( \dfrac{1}{16} \right)-4=0$
Now taking the constant terms on RHS and simplifying we get,
$\begin{align}
  & \Rightarrow {{\left( x-\dfrac{1}{4} \right)}^{2}}=\left( \dfrac{1}{16} \right)+4 \\
 & \Rightarrow {{\left( x-\dfrac{1}{4} \right)}^{2}}=\dfrac{1}{16}+\dfrac{64}{16} \\
 & \Rightarrow {{\left( x-\dfrac{1}{4} \right)}^{2}}=\dfrac{65}{16} \\
\end{align}$
Now taking square root on both sides we get,
$\Rightarrow \left( x-\dfrac{1}{4} \right)=\dfrac{\pm \sqrt{65}}{4}$
Now taking $\dfrac{1}{4}$ to RHS we get,
$\begin{align}
  & \Rightarrow x=\dfrac{1}{4}\pm \dfrac{\sqrt{65}}{4} \\
 & \Rightarrow x=\dfrac{1\pm \sqrt{65}}{4} \\
\end{align}$

Hence the solution of the given equation is $x=\dfrac{1-\sqrt{65}}{4},x=\dfrac{1+\sqrt{65}}{4}$ .

Note: Now note that we can also solve the quadratic equation using the formula. We know that for any quadratic equation of the form $a{{x}^{2}}+bx+c=0$ the solution of the equation is given by $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ . Hence substituting the values of a, b and c we will find the value of x.