Question

By the use of completing square method.Find the roots of equation 2x^2+x+4=0

Answer 1

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

Given quadratic equation is

$\rm :\longmapsto\: {2x}^{2} + x + 4 = 0$

can be rewritten as

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

Divide both sides by 2, to make out the coefficient of x² unity, we get

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

can be rewritten as on adding both sides the square of half the coefficient of x,

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

can be rewritten as

$\rm :\longmapsto\: {x}^{2} + 2 \times \dfrac{1}{4} \times x+ {\bigg[\dfrac{1}{4} \bigg]}^{2} = - \: 2 + \dfrac{1}{16}$

We know,

$\rm :\longmapsto\:\boxed{ \tt{ \: {x}^{2} + 2xy + {y}^{2} = {(x + y)}^{2} \: }}$

So, using this identity, we get

$\rm :\longmapsto\: {\bigg[x + \dfrac{1}{4} \bigg]}^{2} = \dfrac{ - 32 + 1}{16}$

$\rm :\longmapsto\: {\bigg[x + \dfrac{1}{4} \bigg]}^{2} = - \: \dfrac{ 31}{16}$

$\bf\implies \:no \: real \: root \: exist$

More to know :-

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

Answer 2

Answer:

Completing Square Method

Step-by-step explanation:

on dividing 2 we get,

${x}^{2} + \frac{x}{2} - 2 = 0$

Adding and subtracting

$\binom{1}{4} {}^{2} \: we \: get$

${x}^{2} + 2x \binom{1}{4} + \binom{1}{4} {}^{2} - { \binom{1}{4 \:} }^{2} \: - 2 = 0$

$or \: \: (x + \frac{1}{4}) {}^{2} - ( \frac{1}{16} + 2) = 0$

$or \: \: (x + \frac{1}{4} ) {}^{2} - ( { \frac{1 + 32}{16} }) = 0$

$or \: \: (x + \frac{1}{4}) {}^{2} - \frac{33}{16} = 0$

$or \: \: (x + \frac{1}{4} ) {}^{2} = \frac{33}{16}$

(x + 1/4) = ±√33/4

.

. . Roots are x =

$\frac{ - 1 + \sqrt{33} }{4} - ( \frac{1 + \sqrt{33} }{4})$