Question

solve this please fast

Answer 1

$<b>°$

Topic:- Quardratic equation

For real and Equal roots..

we must have:-

D=0

ACCORDING TO THE QUESTION:-


Equation:-

x²-2kx+(7k-12)


D=O

b²-4ac=0

=] (-2k)²-4.1.(7k-12)=)0

=] 4k²-4(7k-12)=)0

=] 4k²-28k+48=0

$<b>[Taking CommOn 4 from Equation]</b>$

=] 4(k²-7k+12)=0

=] k²-7k+12=0

=] k²-4k-3k+12=0

=] k(k-4)-3(k-4)=0

=] (k-4)(k-3)

K=4
K=3

Thanks!!!

Answer 2

Given Equation is x^2 - 2kx + (7k - 12) = 0.

Here a = 1, b = -2k, c = (7k - 12)

Given that the equation has real roots.

⇒ b^2 - 4ac = 0

⇒ (-2k)^2 - 4(1)(7k - 12) = 0

⇒ 4k^2 - 4(7k - 12) = 0

⇒ 4k^2 - 28k + 48 = 0

⇒ k^2 - 7k + 12 = 0

⇒ k^2 - 4k - 3k + 12 = 0

⇒ k(k - 4) - 3(k - 4) = 0

⇒ (k - 3)(k - 4) = 0

⇒ k = 3,4.

Hope this helps!