Question

Find the roots of the following quadratic equations, if they exist, by the method of completing the square :
(i). $2{{x}^{2}}-7x+3=0$
(ii). $2{{x}^{2}}+x-4=0$
(iii). $4{{x}^{2}}+4\sqrt{3}x+3=0$
(iv). $2{{x}^{2}}+x+4=0$

Answer 1

We check the existence of roots of quadratic equation $a{{x}^{2}}+bx+c=0$ by checking whether the discriminant $D={{b}^{2}}-4ac\ge 0$. We use the competing square method first dividing the equation both side by $a$ and then adding ${{\left( \dfrac{-b}{2a} \right)}^{2}}$ both side. We then try to make a whole square in the left hand side using the identity ${{\left( a\pm b \right)}^{2}}={{a}^{2}}+{{b}^{2}}\pm 2ab$$$$$

Complete step-by-step solution:
We know that the quadratic equation in one variable $x$ is given by $a{{x}^{2}}+bx+c=0$ where $a\ne 0,b,c$ are real numbers. The roots for the quadratic equation exists when the discriminant $D={{b}^{2}}-4ac\ge 0$. So let us check whether the given equation has a real root and the use of completing square method to solve. We solve by the completing square method first dividing the equation by $a$ and then add ${{\left( \dfrac{-b}{2a} \right)}^{2}}$ both side of the equation to get a complete square at the left side.
(i) the given quadratic equation is $2{{x}^{2}}-7x+3=0$. Here the discriminant is $D={{\left( -7 \right)}^{2}}-4.2.3=49-24=25>0$. So roots exist. We first divide the terms in both side of the given equation by 2 and then add ${{\left( \dfrac{-7}{2\times 2} \right)}^{2}}={{\left( \dfrac{7}{4} \right)}^{2}}$ both to get

& 2{{x}^{2}}-7x+3=0 \\\ & \Rightarrow {{x}^{2}}-\dfrac{7}{2}x+\dfrac{3}{2}=0 \\\ & \Rightarrow {{x}^{2}}-\dfrac{7}{2}x+\dfrac{3}{2}+{{\left( \dfrac{7}{4} \right)}^{2}}={{\left( \dfrac{7}{4} \right)}^{2}} \\\ & \Rightarrow {{\left( x \right)}^{2}}-2\times x\times \dfrac{7}{4}+{{\left( \dfrac{7}{4} \right)}^{2}}+\dfrac{3}{2}={{\left( \dfrac{7}{4} \right)}^{2}} \\\ \end{aligned} $$ Let us use the algebraic identity of ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$ and proceed $$\begin{aligned} & \Rightarrow {{\left( x \right)}^{2}}-2\times x\times \dfrac{7}{4}+{{\left( \dfrac{7}{4} \right)}^{2}}+\dfrac{3}{2}={{\left( \dfrac{7}{4} \right)}^{2}} \\\ & \Rightarrow {{\left( x-\dfrac{7}{4} \right)}^{2}}={{\left( \dfrac{7}{4} \right)}^{2}}-\dfrac{3}{2}=\dfrac{25}{16} \\\ \end{aligned}$$ We take the square root both side and get, $$\begin{aligned} & \Rightarrow x-\dfrac{7}{4}=\pm \dfrac{5}{4} \\\ & \Rightarrow x=\dfrac{5}{4}+\dfrac{7}{4},x=\dfrac{-5}{4}+\dfrac{7}{4} \\\ & \Rightarrow x=3,x=\dfrac{1}{2} \\\ \end{aligned}$$ (ii) The given quadratic equation is $2{{x}^{2}}+x-4=0$. Here the discriminant is $D={{\left( 1 \right)}^{2}}-4.2.\left( -4 \right)=33>0$. So roots exist. We first divide the terms in both side of the given equation by 2 and then add ${{\left( \dfrac{-1}{2\times 2} \right)}^{2}}={{\left( \dfrac{1}{4} \right)}^{2}}$ both to get $$\begin{aligned} & 2{{x}^{2}}+x-4=0 \\\ & \Rightarrow {{x}^{2}}+\dfrac{1}{2}x-2=0 \\\ & \Rightarrow {{x}^{2}}+\dfrac{1}{2}x-2+{{\left( \dfrac{1}{4} \right)}^{2}}={{\left( \dfrac{1}{4} \right)}^{2}} \\\ & \Rightarrow {{\left( x \right)}^{2}}+2\times x\times \dfrac{1}{4}+{{\left( \dfrac{1}{4} \right)}^{2}}-2={{\left( \dfrac{1}{4} \right)}^{2}} \\\ \end{aligned} $$ Let us use the algebraic identity of ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ and proceed $$\begin{aligned} & \Rightarrow {{\left( x \right)}^{2}}-2\times x\times \dfrac{1}{4}+{{\left( \dfrac{1}{4} \right)}^{2}}-2={{\left( \dfrac{1}{4} \right)}^{2}} \\\ & \Rightarrow {{\left( x-\dfrac{1}{4} \right)}^{2}}={{\left( \dfrac{1}{4} \right)}^{2}}+2=\dfrac{33}{16} \\\ \end{aligned}$$ We take the square root both side and get, $$\begin{aligned} & \Rightarrow x+\dfrac{1}{4}=\pm \dfrac{\sqrt{33}}{4} \\\ & \Rightarrow x=\dfrac{\sqrt{33}}{4}-\dfrac{1}{4},x=\dfrac{-\sqrt{33}}{4}-\dfrac{1}{4} \\\ & \Rightarrow x=\dfrac{\sqrt{33}-1}{4}x=\dfrac{-\sqrt{33}-1}{4} \\\ \end{aligned}$$ (iii) The given quadratic equation is $4{{x}^{2}}+4\sqrt{3}x+3=0$. Here the discriminant is $D={{\left( 4\sqrt{3} \right)}^{2}}-4.4.3=0$. So roots exist. We first divide the terms in both side of the given equation by 4 and then add ${{\left( \dfrac{-4\sqrt{3}}{4\times 2} \right)}^{2}}={{\left( \dfrac{-\sqrt{3}}{2} \right)}^{2}}={{\left( \dfrac{\sqrt{3}}{2} \right)}^{2}}$ both to get $$\begin{aligned} & 4{{x}^{2}}+4\sqrt{3}x+3=0 \\\ & \Rightarrow {{x}^{2}}+{\sqrt{3}}x+\dfrac{3}{4}=0 \\\ & \Rightarrow {{x}^{2}}+{\sqrt{3}}x+\dfrac{3}{4}+{{\left( \dfrac{\sqrt{3}}{2} \right)}^{2}}={{\left( \dfrac{\sqrt{3}}{2} \right)}^{2}} \\\ & \Rightarrow {{\left( x \right)}^{2}}+2\times x\times \dfrac{\sqrt{3}}{2}+{{\left( \dfrac{\sqrt{3}}{2} \right)}^{2}}+\dfrac{3}{4}={{\left( \dfrac{\sqrt{3}}{2} \right)}^{2}} \\\ \end{aligned}$$ Let us use the algebraic identity of ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ and proceed $$\begin{aligned} & \Rightarrow {{\left( x+\dfrac{\sqrt{3}}{2} \right)}^{2}}+\dfrac{3}{4}={{\left( \dfrac{\sqrt{3}}{2} \right)}^{2}} \\\ & \Rightarrow {{\left( x+\dfrac{\sqrt{3}}{2} \right)}^{2}}={{\left( \dfrac{\sqrt{3}}{2} \right)}^{2}}-\dfrac{3}{4}=0 \\\ \end{aligned}$$ We take the square root both side and get, $$ x=\dfrac{-\sqrt{3}}{2}$$ (iv) The given quadratic equation is $2{{x}^{2}}+x+4=0$. Here the discriminant is $D={{\left( 1 \right)}^{2}}-4.2.1=-7< 0$. So roots do not exist.$$$$ **Note:** We note that the roots equal when $D=0$ , the roots are rational when $D$ is a perfect square, the roots are integral when $D$ is a perfect square and $2a$ divides $D$ exactly. We can directly find the roots using the formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.