Question

find the value of k such as kx^2-7x+2=0 has equal roots​

Answer 1

_bjsbs n$: nn n_ znz;z_. Sjsnnsnsb#

Answer 2

,kx2−5x+k=0

Here a=k,b=−5,c=k

Given that the equation has equa roots.

Therefore,

D=0

b2−4ac=0

(−5)−4(k)(k)=0

25−4k2=0

⇒4k2=25

⇒k=±425=±25

Hence the value of k is ±25.