Question

find the value of k such that the following quadratic equation has equal roots.(k-12)x^2-2(k-12)x+2=0

Answer 1

If the quadratic equation has equal roots
so Discriminant =0
d=b^2-4ac
0=(2(k-12))^2-4*(k-12)*2
0=4(k^2+144-24k)-8(k-12)
0=4k^2+576-96k-8k+96
0=4k^2+672-104k
0=4(k^2+168-26k)
0=k^2-26k+168
0=k^2-(14+12)k+168
0=k^2-14k-12k+168
0=k(k-14)-12(k-14)
0=(k-12)(k-14)
k=12,k=14
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Answer 2

$\Large{\textbf{\underline{\underline{According\:to\:the\:Question}}}}$

Equal roots

(k - 12)x² −2(k - 12)x + 2=0.

Situation

If roots are equal then b² = 4ac

Equation :-

(k - 12)x² - 2(k - 12)x+  2

a = k - 12

b = -2(k - 12)

c = 2

Hence here,

$\tt{\rightarrow {-2(k-12)}^2=4\times (k-12)\times 2}$

⇒ 4(k² + 144 - 24k) = 8k - 96

⇒ k² + 144 - 24k = 2k - 24

⇒ k² - 26k + 168 = 0

$k=\frac{26\pm\sqrt{(26)^2-(4\times 168)} }{2}$

$\tt{\rightarrow k=\dfrac{26\pm \sqrt{(676-672)}}{2}}$

$\tt{\rightarrow k=\dfrac{(26\pm \sqrt{4}}{2}}$

$\tt{\rightarrow k=\dfrac{26\pm 2}{2}}$

Hence we get,

⇒ k =  13 ±  1

$\Large{\boxed{\sf\:{Here\;we\;have\;to\;factorise :-}}}$

⇒ k² - 26k + 168 = 0

⇒ k² - 14k - 12k + 168 = 0

⇒ k(k - 14) -12(k - 14) = 0

⇒ (k - 14)(k - 12) = 0

⇒ k = 14 , k = 12

$\Large{\boxed{\sf\:{k = 14\;is\;acceptable\;as\;k = 12\; not\;possible}}}$