Question

For what values of k will be the following equation have real and equal roots: 4x^2-2(k+1)x+(k+4)
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Answer 1

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

The equation

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

have real and equal roots.

$\large\underline{\sf{To\:Find - }}$

The value of k.

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

Given quadratic equation is

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

We know,

A quadratic equation ax² + bx + c = 0, have real and equal roots iff Discriminant, D = 0

So, Let first evaluate Discriminant.

On comparing the given equation with quadratic equation ax² + bx + c = 0, we get

$\red{\rm :\longmapsto\:a = 4}$

$\red{\rm :\longmapsto\:b = - 2(k + 1)}$

$\red{\rm :\longmapsto\:c = k +4}$

So,

$\rm :\longmapsto\:Discriminant, \: D = {b}^{2} - 4ac$

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

$\rm \: = \: \: 4 {(k + 1)}^{2} - 16(k + 4)$

$\rm \: = \: \: 4 {(k}^{2} + 1 + 2k) - 16(k + 4)$

$\rm \: = \: \: 4 {(k}^{2} + 1 + 2k - 4k - 16)$

$\rm \: = \: \: 4 {(k}^{2} - 2k - 15)$

$\rm \: = \: \: 4 {(k}^{2} - 5k + 3k- 15)$

$\rm \: = \: \: 4\bigg(k(k - 5) + 3(k - 5)\bigg)$

$\rm \: = \: \: 4\bigg((k - 5) (k+ 3)\bigg)$

Since, Given equation have real and equal roots.

So,

$\rm :\longmapsto\:D = 0$

$\rm :\longmapsto\:4(k - 5)(k + 3) = 0$

$\rm :\implies\:k - 5 = 0 \: \: or \: \: k + 3 = 0$

$\bf\implies \:k = 5 \: \: or \: \: k = - 3$

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