Find value of K for which …
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Sia ? 6 years, 3 months ago
The quadratic equation {tex}(k-5)x^2 + 2(k-5)x + 2 = 0{/tex} have equal roots.
⇒ Discriminant (b2 - 4ac) = 0
⇒ {tex}[2(k-5)]^2 - 4(k-5)(2) = 0{/tex}
⇒ {tex}4(k^2 - 10k + 25) - (8k - 40) = 0{/tex}
⇒ {tex}4k^2 - 40k + 100 - 8k + 40 = 0{/tex}
⇒ 4k2 - 48k + 140 = 0
⇒ k2 - 12k + 35 = 0
⇒ (k - 7)(k - 5) = 0
⇒ k = 7 or 5
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