in the equation x²+2(k+2)x +9k =0 has equal roots, find k
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Hey dear friend,
Equation=x²+2(k+2)x+9k=0
Compare with ax²+bx+c=0
a=1, b=2(k+2)=>(2k+4) and c=9k
By discriminant formula =>b²-4ac
Substituting values
=> (2k+4)²- 4*1*9k=0
=> 4k²+16+16k-36k=0
=>4k²-20k+16=0
Solving by factorisation
=>4k²-16k-4k+16=0
=>4k(k-4) -4(k-4)=0
=>(4k-4)(k-4)=0
4k-4=0. k-4=0
4k=4. k=4
k=1
So, the value of k can be 1 or 4.
perfectly fine pure correct answer
hope it helps you mark me as brainliest and follow me
Equation=x²+2(k+2)x+9k=0
Compare with ax²+bx+c=0
a=1, b=2(k+2)=>(2k+4) and c=9k
By discriminant formula =>b²-4ac
Substituting values
=> (2k+4)²- 4*1*9k=0
=> 4k²+16+16k-36k=0
=>4k²-20k+16=0
Solving by factorisation
=>4k²-16k-4k+16=0
=>4k(k-4) -4(k-4)=0
=>(4k-4)(k-4)=0
4k-4=0. k-4=0
4k=4. k=4
k=1
So, the value of k can be 1 or 4.
perfectly fine pure correct answer
hope it helps you mark me as brainliest and follow me
PranavMudgal:
bro if you have method then answer int his questions
the answer is correct
PranavMudgal
follow me athul
I have one more question
i have posted my answer
please check it and tell me wether it is correct or not mr. professor
correct bro
thank you
at last you agreed
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this method is easier than the other one that mr. Pranav has done
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