Question

if the roots of the equation x2+x+a=0 be real and unequal ,then prove that the roots of the equation 2x^2 -4(1+a)x+2a^2+3=0 are imaginary (a is real)

Answer 1

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

Concept Used :-

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

Now,

Here, it is given that

$\rm :\longmapsto\: {x}^{2} + x + a = 0\: has \: real \: and \: unequal \: roots.$

$\rm :\implies\:Discriminant, \: D > 0$

$\rm :\implies\: {b}^{2} - 4ac > 0$

Here, on comparing with ax² + bx + c = 0, we get

$\rm :\longmapsto\:a = 1$

$\rm :\longmapsto\:b = 1$

$\rm :\longmapsto\:c = a$

So, on substituting the values in above, we get

$\rm :\longmapsto\: {(1)}^{2} - 4a > 0$

$\rm :\longmapsto\: 1 - 4a > 0$

$\bf\implies \:4a - 1 < 0 - - - (1)$

Now,

Consider,

$\rm :\longmapsto\: {2x}^{2} - 4(1 + a) + {2a}^{2} - 3 = 0$

Now, further Consider,

Discriminant,

$\rm :\longmapsto\:Discriminant, \:D = {B}^{2} - 4AC$

Here, on comparing with Ax² + Bx + C = 0 we get

$\rm :\longmapsto\:A = 2$

$\rm :\longmapsto\:B = - 4(1 + a)$

$\rm :\longmapsto\:C = 2 {a}^{2} - 3$

On substituting the values, we get

$\rm \: = \: \: {(4(1 + a))}^{2} - 4 \times 2 \times ( {2a}^{2} + 3)$

$\rm \: = \: \:16(1 + {a}^{2} + 2a) - 8( {2a}^{2} + 3)$

$\rm \: = \: \:8\bigg(2(1 + {a}^{2} + 2a) - ( {2a}^{2} + 3) \bigg)$

$\rm \: = \: \:8\bigg(2 + 2{a}^{2} + 4a - {2a}^{2} - 3 \bigg)$

$\rm \: = \: \:8\bigg( 4a - 1\bigg)$

$\bf\implies \:Discriminant, \: D < 0$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \red{\bigg \{ \because \:4a \: - \: 1 \: < \: 0 \bigg \}}$

Hence, Roots of the equation are imaginary.