Question

2x^2-2x+1/2=0 by perfect square method​

Answer 1

Given Question :-

Solve the following equation by perfect square method :-

$\rm :\longmapsto\: {2x}^{2} - 2x + \dfrac{1}{2} = 0$

$\large\underline{\bf{Solution-}}$

$\rm :\longmapsto\: {2x}^{2} - 2x + \dfrac{1}{2} = 0$

$\rm :\longmapsto\: {2x}^{2} - 2x = - \: \dfrac{1}{2}$

Step :- 1 Make the coefficient of x² unity.

So, Divide both sides by 2.

$\rm :\longmapsto\: {x}^{2} - x = - \: \dfrac{1}{4}$

Step :- 2 Adding the square of half the coefficient of x on both sides.

We get

$\rm :\longmapsto\: {x}^{2} - x + {\bigg( - \dfrac{1}{2} \bigg) }^{2} = - \: \dfrac{1}{4} + {\bigg( - \dfrac{1}{2} \bigg) }^{2}$

$\rm :\longmapsto\: {x}^{2} - 2 \times x \times \dfrac{1}{2} + {\bigg(\dfrac{1}{2} \bigg) }^{2} = - \: \dfrac{1}{4} + {\bigg(\dfrac{1}{4} \bigg) }$

$\rm :\longmapsto\: {\bigg(x - \dfrac{1}{2} \bigg) }^{2} = 0$

$\: \: \: \: \: \: \: \: \: \: \: \red{\boxed{ \bf \: \because \: {x}^{2} - 2xy + {y}^{2} = {(x - y)}^{2}}}$

$\rm :\longmapsto\:{\bigg(x - \dfrac{1}{2} \bigg) } = 0$

$\bf\implies \:x = \dfrac{1}{2}$

Additional Information :-

Nature of roots :-

Let us consider a quadratic equation ax² + bx + c = 0, then nature of roots of quadratic equation depends upon Discriminant (D) of the quadratic equation.

• If Discriminant, D > 0, then roots of the equation are real and unequal.

• If Discriminant, D = 0, then roots of the equation are real and equal.

• If Discriminant, D < 0, then roots of the equation are unreal or complex or imaginary.

Where,

• Discriminant, D = b² - 4ac