Question

Find m if
(m - 12) x ²+ 2(m-12)x +2 = 0
has real roots and equal​

Answer 1

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

$\sf \: A \: quadratic \: equation \: (m - 12) {x}^{2} + 2(m - 12)x + 12 = 0$

$\sf \: has \: real \: and \: equal \: roots.$

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

$\sf \: The \: value \: of \: m$

$\large\underline{\sf{Basic \: Concept \: Used - }}$

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

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

Given

$\sf \: A \: quadratic \: eq^{n} \: {(m - 12)x}^{2} + 2(m - 12)x + 2 = 0 - - (1)$

On comparing with ax² + bx + c = 0, we get

• a = m - 12

• b = 2(m - 12)

• c = 2

So,

Discriminant (D) of above quadratic equation is given by

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

$\rm :\longmapsto\:D = { \{2(m - 12) \}}^{2} - 4 \times 2 \times (m - 12)$

$\rm :\longmapsto\:D = 4 {(m - 12)}^{2} - 8(m - 12)$

$\rm :\longmapsto\:D = 4(m - 12)(m - 12 - 2)$

$\bf\implies \:D = 4(m - 12(m - 14)$

Now,

It is given that,

• Quadratic equation (1) has real and equal roots,

$\bf\implies \:Discriminant \: (D) = 0$

$\rm :\longmapsto\:4(m - 12)(m - 14) = 0$

$\rm :\implies\:m \: = 14 \: and \: m = 12 \: \{rejected \because \: {eq}^{n} \: not \: exist \}$

$\bf\implies \:m \: = \: 14$

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

$\sf \: If \: m = 12, \: then \: equation \: (1) \: can \: be \: rewritten \: as$

$\rm :\longmapsto\:\sf \: (12 - 12) {x}^{2} + 2(12 - 12)x + 2 = 0$

$\rm :\longmapsto\:0 + 0 + 2 = 0$

$\rm :\longmapsto\:2 = 0$

$\rm :\longmapsto\:which \: is \: absurd.$