solve this please fast
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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
=] 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!!!
Answered by
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!
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