Question

find the value of a such that the equation
(a - 3) x ² + 4 (a-3) x + 4 = 0
has equal roots,​

Answer 1

Question:

Find the value of a such that the equation (a-3)x^2 + 4(a-3)x + 4 = 0

has equal roots.

Solution:

Here ,

The given quadratic equation is;

(a-3)x^2 + 4(a-3)x + 4 = 0.

Thus,

Determinant = B^2 - 4•A•C

=> D = [4(a-3)]^2 - 4•(a-3)•4

=> D = 16(a-3)^2 - 16(a-3)

=> D = 16(a-3){(a-3) - 1}

=> D = 16(a-3)(a-4)

Also,

For the given quadratic equation to have real and equal roots , its determinant must be equal to zero.

ie;

=> D = 0

=> 16(a-3)(a-4) = 0

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

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

=> a = 3 OR a = 4

Hence,

For real and equal roots, "a" can take values of 3 and 4.

Answer 2

Answer:

→a = 3 or 4

Step-by-step explanation:

$\mathfrak{given \: equation \: is }→\\ \\ \bf{ {(a - 3)x}^{2} + 4(a - 3)x + 4 = 0}$

According to the question.→

The equation has equal roots .

Hence ,It's discriminant ( D )=0

we know that,

$Discriminant \: (D) = {B}^{2} - 4AC\\ \\ here \: \\ A =\bf{ (a - 3) \: ,\: B= 4(a - 3) \:, \: C = 4 }\\ \\ \therefore \: D = \bf{{ \bigg(4(a - 3) \bigg)}^{2} - 4 \times (a - 3) \times 4}$

$\because \: D = 0 \\ \\ \bf{0 = 16( {a}^{2} + 9 - 2 \times 3 \times a) - 16(a - 3)}$

$0 = \bf{16 {a}^{2} + 144 - 96a - 16a + 48 }\\ \\ \bf{0 = 16 {a}^{2} - 112a + 192}$

$0 = \bf{16 {a}^{2} + 144 - 96a - 16a + 48 }\\ \\ \bf{0 = 16 {a}^{2} - 112a + 192} \\ \\ \bf{ 0 = {a}^{2} - 7a + 12} \\ \\ \bf{0 = {a}^{2} - 4a - 3a + 12} \\ \\ \bf{ 0 = a(a - 4) - 3(a - 4) }\\ \\ \bf{ 0 = (a - 3)(a - 4)} \\ \\ \bf{if \: (a - 3) = 0 }\: \\ \bf{a = 3} \\ \\ \bf{if \: (a - 4) = 0 }\\ \bf{ a = 4} \\ \\ \bf{a = 3 \: or \: 4} \\ \\ \mathfrak{hope \: it \: helps \: you}$