Question

Given the quadriatic equation

x² + 6x + 9 = 4kx
find the set values for which k has real roots​

Answer 1

⇒ The given quadratic equation is x2−kx+9=0, comparing it with ax2+bx+c=0

∴ We get, a=1b=−k,c=9

⇒ It is given that roots are real and distinct.

∴ b2−4ac>0

⇒ (−k)2−4(1)(9)>0

⇒ k2−36>0

⇒ k2>36

⇒ k>6 or k<−6

∴ We can see values of k given in question are correct.