>
Find the roots of the following quadratic equations, if they exist, by the method of completing the square:iv 2 x2+x+4=0
Suggest Corrections
0
Quadratic is a type of problem which deals with a variable multiplied by itself- an operation also known as squaring.
Solution:
Given equation,
${{x}^{2}}+\frac{11}{3}+\frac{10}{3}=0$
${{x}^{2}}+2\times \frac{1}{2}\times \frac{11x}{3}+\frac{10}{3}=0$
${{x}^{2}}+2\times \frac{11x}{6}+{{\left( \frac{11}{6} \right)}^{2}}-{{\left( \frac{11}{6} \right)}^{2}}+\frac{10}{3}=0$
${{\left( x+\frac{11}{6} \right)}^{2}}={{\left( \frac{11}{6} \right)}^{2}}-\frac{10}{3}$
${{\left( x+\frac{11}{6} \right)}^{2}}=\frac{121}{36}-\frac{10}{3}$
${{\left( x+\frac{11}{6} \right)}^{2}}=\frac{121-120}{36}$
${{\left( x+\frac{11}{6} \right)}^{2}}=\frac{1}{36}$
${{\left( x+\frac{11}{6} \right)}^{2}}={{\left( \frac{1}{6} \right)}^{2}}$
$x+\frac{11}{6}=\pm \frac{1}{6}$
$x+\frac{11}{6}=\frac{1}{6}orx+\frac{11}{6}=\frac{-1}{6}$
$x=\frac{1}{6}-\frac{11}{6}orx=\frac{-1}{6}-\frac{11}{6}$
$x=\frac{-10}{6}orx=\frac{-12}{6}=-2$
${{x}^{2}}+\frac{11x}{3}+\frac{10}{3}=0$
${{x}^{2}}+2\times \frac{1}{2}\times \frac{11x}{3}+\frac{10}{3}=0$
${{x}^{2}}+2\times \frac{11x}{6}+{{\left( \frac{11}{6} \right)}^{2}}-{{\left( \frac{11}{6} \right)}^{2}}+\frac{10}{3}=0$
${{\left( x+\frac{11}{6} \right)}^{2}}={{\left( \frac{11}{6} \right)}^{2}}-\frac{10}{3}$
${{\left( x+\frac{11}{6} \right)}^{2}}=\frac{121}{36}-\frac{10}{3}$
${{\left( x+\frac{11}{6} \right)}^{2}}=\frac{1}{36}$
${{\left( x+\frac{11}{6} \right)}^{2}}={{\left( \frac{1}{6} \right)}^{2}}$
$x+\frac{11}{6}=\pm \frac{1}{6}$
$x+\frac{11}{6}=\frac{1}{6}orx+\frac{11}{6}=\frac{-1}{6}$
$x=\frac{1}{6}-\frac{11}{6}orx=\frac{-1}{6}-\frac{11}{6}$
$x=\frac{-10}{6}oex=\frac{-12}{6}=-2$
⇒ $x=-5/3=3$ or $x=-2$
Thus, the roots of the given quadratic equation are $x=-5/3$ and $x=-2.$
Solutions: $$ \begin{array}{l} \text { (i) } 2 x^{2}-7 x+3=0 \\ \Rightarrow 2 x^{2}-7 x=-3 \end{array} $$ Dividing by 2 on both sides, we get $$ \begin{array}{l} \Rightarrow x^{2}-7 x / 2=-3 / 2 \\ \Rightarrow x^{2}-2 \times x \times 7 / 4=-3 / 2 \end{array} $$ On adding $(7 / 4)^{2}$ to both sides of equation, we get $$ \begin{array}{l} \Rightarrow(x)^{2}-2 x x \times 7 / 4+(7 / 4)^{2}=(7 / 4)^{2}-3 / 2 \\ \Rightarrow(x-7 / 4)^{2}=(49 / 16)-(3 / 2) \\ \Rightarrow(x-7 / 4)^{2}=25 / 16 \\ \Rightarrow(x-7 / 4)^{2}=\pm 5 / 4 \\ \Rightarrow x=7 / 4 \pm 5 / 4 \\ \Rightarrow x=7 / 4+5 / 4 \text { or } x=7 / 4-5 / 4 \\ \Rightarrow x=12 / 4 \text { or } x=2 / 4 \\ \Rightarrow x=3 \text { or } x=1 / 2 \end{array} $$ (ii) $2 x^{2}+x-4=0$ $\Rightarrow 2 x^{2}+x=4$ Dividing both sides of the equation by 2 , we get $$ \Rightarrow x^{2}+x / 2=2 $$ Now on adding $(1 / 4)^{2}$ to both sides of the equation, we get, $$ \begin{array}{l} \Rightarrow(x)^{2}+2 \times x \times 1 / 4+(1 / 4)^{2}=2+(1 / 4)^{2} \\ \Rightarrow(x+1 / 4)^{2}=33 / 16 \\ \Rightarrow x+1 / 4=\pm \sqrt{3} 3 / 4 \\ \Rightarrow x=\pm \sqrt{33 / 4}-1 / 4 \\ \Rightarrow x=\pm(\sqrt{33-1}) / 4 \end{array} $$ Therefore, either $x=(\sqrt{33}-1) / 4$ or $x=(-\sqrt{33}-1) / 4$ (iii) $4 x^{2}+4 \sqrt{3} x+3=0$ Converting the equation into $a^{2}+2 a b+b^{2}$ form, we get, $$ \begin{array}{l} \Rightarrow(2 x)^{2}+2 \times 2 x \times \sqrt{3}+(\sqrt{3})^{2}=0 \\ \Rightarrow(2 x+\sqrt{3})^{2}=0 \\ \Rightarrow(2 x+\sqrt{3})=0 \text { and }(2 x+\sqrt{3})=0 \end{array} $$ Therefore, either $x=-\sqrt{3} / 2$ or $x=-\sqrt{3} / 2$. $$ \begin{array}{l} \text { (iv) } 2 x^{2}+x+4=0 \\ \Rightarrow 2 x^{2}+x=-4 \end{array} $$ Dividing both sides of the equation by 2 , we get $$ \begin{array}{l} \Rightarrow x^{2}+1 / 2 x=2 \\ \Rightarrow x^{2}+2 \times x \times 1 / 4=-2 \end{array} $$ By adding $(1 / 4)^{2}$ to both sides of the equation, we get $$ \begin{array}{l} \Rightarrow(x)^{2}+2 \times x \times 1 / 4+(1 / 4)^{2}=(1 / 4)^{2}-2 \\ \Rightarrow(x+1 / 4)^{2}=1 / 16-2 \\ \Rightarrow(x+1 / 4)^{2}=-31 / 16 \end{array} $$ As we know, the square of numbers cannot be negative.
Therefore, there is no real root for the given equation, $2 x^{2}+x+4=0$.