Question

Solve‌ ‌the‌ ‌quadratic‌ ‌equation‌ ‌$4{{x}^{2}}-8x+1=0$‌ ‌?

Answer 1

Hint: We use the completing square method to solve the quadratic equation $a{{x}^{2}}+bx+c=0$. We first take $c$ to the right hand side of the equation then $a\ne 1$ we divide both sides by $a$. We then add ${{\left( \dfrac{-b}{2a} \right)}^{2}}$ both sides. We make a complete square and then take the square root on both sides.

Complete step-by-step solution:
We are given the quadratic equation $4{{x}^{2}}-8x+1=0$ in the question; we compare it with general quadratic equation $a{{x}^{2}}+bx+c=0$ to find $a=4,b=-8,c=1$. We follows steps of completing the square method and take $c$ to the right hand side to have
\[ 4{{x}^{2}}-8x=-1\]
Since $a=4\ne 1$ we divide both sides of the above equation by 4 to have;
\[ \Rightarrow {{x}^{2}}-2x=-\dfrac{1}{4} \]
We add ${{\left( \dfrac{-b}{2a} \right)}^{2}}={{\left( \dfrac{-\left( -8 \right)}{2\times 4} \right)}^{2}}={{\left( \dfrac{8}{8} \right)}^{2}}={{1}^{2}}=1$ both side sides of the above equation to have
\[ \Rightarrow {{x}^{2}}-2x+1=-\dfrac{1}{4}+1\]Now we shall make a complete square using the terms in the left hand side of the above step. Let us have;
\[ \Rightarrow {{\left( x \right)}^{2}}-2\times x\times 1+{{\left( 1 \right)}^{2}}=\dfrac{-1+4}{4}\]
We use the algebraic identity ${{a}^{2}}-2ab+{{b}^{2}}={{\left( a-b \right)}^{2}}$ in the left hand side of the above step to have the complete square as
\[\Rightarrow {{\left( x-1 \right)}^{2}}=\dfrac{3}{4}\]
 We take square root both sides of the above step to have;
\[\begin{align}
  & \Rightarrow \left( x-1 \right)=\pm \dfrac{\sqrt{3}}{2} \\
 & \Rightarrow x=1\pm \dfrac{\sqrt{3}}{2} \\
 & \Rightarrow x=1+\dfrac{\sqrt{3}}{2},1-\dfrac{\sqrt{3}}{2} \\
\end{align}\]
So we got two irrational roots of the quadratic equation as $x=1+\dfrac{\sqrt{3}}{2},1-\dfrac{\sqrt{3}}{2}$.

Note: We note that two irrational numbers $a+\sqrt{b},a-\sqrt{b}$ where $b$ is not a perfect square are called conjugates. The roots of a quadratic equation if they are irrational they are always conjugate of each other for example here in this problem $1+\dfrac{\sqrt{3}}{2},1-\dfrac{\sqrt{3}}{2}$. Another method to solve the quadratic equation is splitting the middle term $b$ but we cannot use it here since the discriminant $D={{b}^{2}}-4ac$ is not a perfect square.