Question

The values of k for which the roots of equation(x-1)(x-5)+k=0 differ by2 is

Answer 1

=> k = 8

given eqn :-

(x-1)(x-5)+k=0

=> x²-6x+k =0

since roots of eqn differ by 2

=> first root = p

second root= p+2

sum of the roots

= p+(p+2) = 6/1 = 6

=> 2p = 4

=> p = 2..............(1)

again, product of the roots

= p(p+2) = k/1

=> 2(2+2) = k

=> k = 2×4

= 8

THUS FOR K=8 ,THE GIVEN EQUATION HAS ROOTS SUCH THAT THESE DIFFER BY 2.

HOPE THIS HELPS YOU

Answer 2

Given(x−5)(x−1)+k=0⇒x2−6x+5+k=0−−−−(1)Itisintheformax2+bx+c=0∴a=1,b=−6,c=(5+k)Ifαandβaretherootsthenα+β=6−−−−(2)αβ=(5+k)−−−−(3)andgivenα−β=2−−−−(4)Adding(2)and(4)weget2α=8⇒α=4−−−−(5)from(3)and(4)4+β=6∴β=2−−−−(6)∴from(3),(5)and(6)weget5+k=4×2⇒k=3(Ans)