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Question: Solve‌ ‌the‌ ‌quadratic‌ ‌equation‌ ‌\(4{{x}^{2}}-8x+1=0\)‌ ‌?...

Solve‌ ‌the‌ ‌quadratic‌ ‌equation‌ ‌4x2−8x+1=04{{x}^{2}}-8x+1=0‌ ‌?

Explanation

Solution

We use the completing square method to solve the quadratic equation ax2+bx+c=0a{{x}^{2}}+bx+c=0. We first take cc to the right hand side of the equation then a≠1a\ne 1 we divide both sides by aa. We then add (−b2a)2{{\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 4x2−8x+1=04{{x}^{2}}-8x+1=0 in the question; we compare it with general quadratic equation ax2+bx+c=0a{{x}^{2}}+bx+c=0 to find a=4,b=−8,c=1a=4,b=-8,c=1. We follows steps of completing the square method and take cc to the right hand side to have
4x2−8x=−14{{x}^{2}}-8x=-1
Since a=4≠1a=4\ne 1 we divide both sides of the above equation by 4 to have;
⇒x2−2x=−14\Rightarrow {{x}^{2}}-2x=-\dfrac{1}{4}
We add (−b2a)2=(−(−8)2×4)2=(88)2=12=1{{\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
⇒x2−2x+1=−14+1\Rightarrow {{x}^{2}}-2x+1=-\dfrac{1}{4}+1Now we shall make a complete square using the terms in the left hand side of the above step. Let us have;
⇒(x)2−2×x×1+(1)2=−1+44\Rightarrow {{\left( x \right)}^{2}}-2\times x\times 1+{{\left( 1 \right)}^{2}}=\dfrac{-1+4}{4}
We use the algebraic identity a2−2ab+b2=(a−b)2{{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
⇒(x−1)2=34\Rightarrow {{\left( x-1 \right)}^{2}}=\dfrac{3}{4}
We take square root both sides of the above step to have;

& \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{aligned}$$ **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.