Question

3) If the roots of the given quadratic equation are real and equal then find
the value of m'.
(m-12) x² + 2 (m-12) x + 2 = 0​

Answer 1

Answer:

14

Step-by-step explanation:

To roots to be real and equal, discriminant of the equation must be 0.

Discriminant of ax² + bx + c = 0 is given by b² - 4ac.   On comparing,

a = (m - 12),   b = 2(m - 12),     c = 2

  ⇒ discriminant = 0

⇒ [2(m - 12)]² - 4(2)(m - 12) = 0

⇒ 4(m - 12)² - 8(m - 12) = 0

⇒ 4(m - 12)[ (m - 12) - 2 ] = 0

⇒ 4(m - 12)(m - 14) = 0

⇒ m - 12 = 0    or   m - 14 = 0

⇒ m = 12      or m = 14

But for m = 12,  (m - 12)x² + 2(m - 12) + 2 = 0  is not true.

m = 14 must be preferred

Answer 2

$\sf Equal \: and \: real \: roots \: ⇒b² - 4ac = 0$

$\sf \therefore b² = 4ac$

$\sf \: a=(m−12),b=2(m−12),c=2$

$\sf [2(m−12)] {}^{2} =4(m−12)(2)$

$\sf4(m−12) {}^{2} =4(m−12)(2)$

$\sf (m−12) {}^{2} −2(m−12)=0$

$\sf(m−12)(m−12−2)=0$

$\sf∴m=14,12$