This is the question...
Attachments:
Answers
Answered by
0
putting values of x in eq 1 we get the value of p as 2.Then putting same value of x and p in eq 2 we get the value of k as -7
Tinakumar:
But its not given that x=-2 is the root of next equation.
yes it is given.it is said that root of the eq should be same.Actually what is the root in the second eq.it is just x and nothing else.
But its given that the two roots of equation x^2+k(4x+k-1)+p=0 are equal. It means that D=b^-4ac=0
yes you are right.k=2/3 or -1
x^2+4kx+k^2-k+p=0.....then rearranging to quadratic eq we get...(x^2)+(4kx)+(k^2-k+p)=0...acc to qstn roots are same...so b^2=4ac...then by putting respective values we get...the answer...thank you...nice qstn...
here a=1,b=4k,c=k^2-k+p...so put in b^2=4ac and get the desired value of k...
Thanks
you are welcome.
Answered by
1
at first u hv to find out the P
3.4+7.-2+p=0
12-14+p=0
p=2
then the other eq
4+-2.(4.-2+k-1)+p=0
4+-2.(k-9)+2=0
4-2k+18+2=0
24=2k
k =12
3.4+7.-2+p=0
12-14+p=0
p=2
then the other eq
4+-2.(4.-2+k-1)+p=0
4+-2.(k-9)+2=0
4-2k+18+2=0
24=2k
k =12
Similar questions