Question

If 3 is a root of the quadratic equation x2-x+k=0 find the vqlue of k for which the quadratic equation x2+px(1+3k)+7(3+2k)=0 has equal roots

Answer 1

Answer:

so if 3 is a root to equation, then it will satisfy the equation...

therefore

(3)^2-3+k=0

9-3+k=0

6+k=0

k= -6

now substituting k in below equation we get

x2 +px(-17) + 7(-9) =0

x2 -17px -63=0

for equal roots discriminate should be zero, therefore

b2-4ac=0

(-17p)^2 - 4*1*(-63) =0

289*p^2 + 252 =0

p^2= -(252/289)

which doesnot exists, so there will be no value of p for which the equation have equal roots