Question

Give the value of 'a' such that (a-3)x² + 4 (a-3) x + 4 = 0 has equal roots.​

Answer 1

Step-by-step explanation:

Let,

p(x) = (a-3)x² + 4(a-3)x + 4 = 0 --------------(1)

Comparing (1) by ax² + bx + c = 0, we get:

a = (a-3)

b = 4(a-3)

c = 4

Since it is given that given quadratic polynomial has equal roots then it's discriminate should be zero.

Therefore,

D = 0

b² - 4ac = 0

{4(a-3)}² - 4(a-3)(4) = 0

{16(a² + 9 - 2(a)(3)} - 16(a-3) = 0

16(a² + 9 - 6a) - 16(a-3) = 0

16(a² + 9 - 6a - a + 3) = 0

a² - 7a + 12 = 0

a² - 4a - 3a + 12 = 0

a(a-4) -3(a-4) = 0

(a-4)(a-3) = 0

Therefore,

a = 4 and a = 3

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