Find the roots of the following quadratic equations (if they exist) by the method of completing the square. $2{{x}^{2}}-7x+3=0$

3 years ago

Quadratic is a type of problem which deals with a variable multiplied by itself- an operation also known as squaring.

Solution:

Given equation,

$2{{x}^{2}}-7x+3=0$

$2\left( {{x}^{2}}-\frac{7x}{2}+\frac{3}{2} \right)=0$

${{x}^{2}}-2\times \frac{1}{2}\times \frac{7}{2}\times x+\frac{3}{2}=0$

${{x}^{2}}-2\times \frac{7}{4}\times x+{{\left( \frac{7}{4} \right)}^{2}}-{{\left( \frac{7}{4} \right)}^{2}}+\frac{3}{2}=0$

${{\left( x-\frac{7}{4} \right)}^{2}}-\frac{49}{16}+\frac{3}{2}=0$

${{\left( x-\frac{7}{4} \right)}^{2}}=\frac{49}{16}-\frac{3}{2}$

${{\left( x-\frac{7}{4} \right)}^{2}}=\frac{49-26}{16}$

${{\left( x-\frac{7}{4} \right)}^{2}}=\frac{25}{16}$

${{\left( x-\frac{7}{4} \right)}^{2}}={{\left( \frac{5}{4} \right)}^{2}}$

$x-\frac{7}{4}=\pm \frac{5}{4}$

$x-\frac{7}{4}=\frac{5}{4}orx-\frac{7}{4}=-\frac{5}{4}$

$x=\frac{7}{4}+\frac{5}{4}orx=\frac{7}{4}-\frac{5}{4}$

$2\left( {{x}^{2}}-\frac{7x}{2}+\frac{3}{2} \right)=0$

${{x}^{2}}-2\times \frac{7}{2}\times \frac{1}{2}\times x+\frac{3}{2}=0$

${{x}^{2}}-2\times \frac{7}{2}\times x+{{\left( \frac{7}{4} \right)}^{2}}-{{\left( \frac{7}{4} \right)}^{2}}+\frac{3}{2}=0$

${{x}^{2}}-2\times \frac{7}{4}\times x+{{\left( \frac{7}{4} \right)}^{2}}-\left( \frac{49}{16} \right)+\frac{3}{2}=0$