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:

(m-12)x² +(m-12)x +2=0

we have to find Discriminant

that is

b²-4ac=0{because roots are real and equal}

(m-12)²-4(m-12)(2)=0

(m-12)²-8(m-12)=0

m²+12²-2(12)(m) -8m+96=0

{(a-b)²=a²+b²-2ab. or a²-2ab+b²}

m²+144-24m-8m+96=0

m²+240-32m=0

m²-32m+240=0

m²-12m-20m+240=0

m(m-12)-20(m-12)=0

(m-12)(m-20)=0

if m-12=0

then m=12

if m-20=0

then

m=20

please mark me as brainlist it is too important for me

Answer 2

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