Find value of K for which the quadratic equation (K-5)x2+2(k-5)x+2=0

1.	Find value of K for which the quadratic equation (K-5)x2+2(k-5)x+2=0
Answer» The quadratic equation {tex}(k-5)x^2\xa0+ 2(k-5)x + 2 = 0{/tex} have equal roots.⇒ Discriminant (b2\xa0- 4ac) = 0⇒ {tex}[2(k-5)]^2\xa0- 4(k-5)(2) = 0{/tex}⇒ {tex}4(k^2\xa0- 10k + 25) - (8k - 40) = 0{/tex}⇒ {tex}4k^2\xa0- 40k + 100 - 8k + 40 = 0{/tex}⇒ 4k2\xa0- 48k + 140 = 0⇒ k2\xa0- 12k + 35 = 0⇒ (k - 7)(k - 5) = 0⇒ k = 7 or 5

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