Question

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:

Solution:-

(m - 12)x² + 2(m - 12) x + 2 = 0

comparing with ax²+bx+c = 0

we get, a = m - 12 , b=2(m - 12) and c= 2

b²-4ac²

= [2(m - 12)]²-4×( m-12)×2

= 4(m-12)² - 8( m-12)

= 4(m²-24m+144)-8m+96

= 4m²-96m+576-8m+96

= 4m²-104m+672

But this equation has real and equal root

since, b²-4ac=0

4m²-104m+672 = 0

4(m²-26m+168) = 0

${m}^{2} - 26m + 168 = \frac{0}{4}$

m² - 26m + 168 = 0

m² - 14m - 12m + 168 = 0

m(m - 14) -12(m - 14) = 0

(m - 14) (m - 12) = 0

m- 14 = 0. or. m - 12 = 0

m = 14. m = 12

but m = 12 doesn't satisfy the equation

since, m = 14