Question

Find the value of alpha for which the equation

$( \alpha - 12)x {}^{2} + 2( \alpha - 12)x + 2 = 0$
has equal roots.

Answer 1

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(a – 12)x2 + 2(a –12)x + 2 = 0

The given equations will have equal roots, if its discriminant is zero, i.e.,

D = b2 – 4ac = 0

Now, from the above mentioned equations, we have

b = 2(a –12)

a = (a – 12)

c = 2

∴ D = b2 – 4ac = 0

⇒ [2(a – 12)]2 – 4 (a – 12) × 2 = 0

⇒ 4(a – 12)2 – 8 (a – 12) = 0

⇒ 4(a – 12) [(a – 12) – 2] = 0

⇒ [a – 14] = 0 or a = 12

⇒ a = 14 or a = 12

The value of a cannot be equal to 12 as for a= 12, the given equation becomes:

(12 – 12)x2 + 2(12 – 12)x + 2 = 0

⇒ ox2 + ox + 2 = 0

which cannot be possible

∴ For a = 14, the given equations has equal 

Answer 2

(a – 12)x2 + 2(a –12)x + 2 = 0

The given equations will have equal roots, if its discriminant is zero, i.e.,

D = b2 – 4ac = 0

Now, from the above mentioned equations, we have

b = 2(a –12)

a = (a – 12)

c = 2

∴ D = b2 – 4ac = 0

⇒ [2(a – 12)]2 – 4 (a – 12) × 2 = 0

⇒ 4(a – 12)2 – 8 (a – 12) = 0

⇒ 4(a – 12) [(a – 12) – 2] = 0

⇒ [a – 14] = 0 or a = 12

⇒ a = 14 or a = 12

The value of a cannot be equal to 12 as for a= 12, the given equation becomes:

(12 – 12)x2 + 2(12 – 12)x + 2 = 0

⇒ ox2 + ox + 2 = 0

which cannot be possible

∴ For a = 14, the given equations has equal