Question

Find the value of quadratic equation (2k+1)square +2(k+3)x+(k+5)=0 has real and equal roots

Answer 1

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Answer 2

GiveN Quad. eq.

(2k+3)x² + 2(k+3)x + (k+5)=0

The equation has equal roots.

To FinD:

Value of k here?

Step-wise-Step Explanation:

When any Quadratic equation has equal roots, the Discriminate is always equals to 0. Now Discriminate means$\rm{ {b}^{2} - 4ac}$

when the equation is compared to$\rm{a {x}^{2} + bx + c}$

Here,

a = 2k + 3

b = 2k + 6

c = k + 5

We know that D = 0. Putting the values of a, b and c from this equation.

⇒ b² - 4ac = 0

⇒ (2k + 6)² - 4(2k + 3)(k + 5) = 0

⇒ 4k² + 24k + 36 - 4(2k² + 13k + 15) = 0

⇒ 4k² + 24k + 36 - 8k² - 52k - 60 = 0

⇒ -4k² - 28k - 24 = 0

⇒ 4k² + 28k + 24 = 0

Now factorising the above quad. eq. using middle term factorisation,

⇒ 4k² + 4k + 24k + 24 = 0

⇒ 4k(k + 1) + 24(k + 1) = 0

⇒ (4k + 24)(k + 1) = 0

⇒ (k + 6)(k + 1) = 0

Equating to 0, we will get k = -6 or -1 (Ans)