Question

For what value of m does the quadratic equation (m-5)x2 + 2(m-5)x + 2 = 0 have equal (2) roots?

Answer 1

Answer:

The given equation is x

2

+2(m−1)x+(m+5)=0

For the equation to have real and equal roots, the discriminant should be 0.

∴[2(m−1)]

2

−4(m+5)=0

⇒4(m

2

−2m+1)=4(m+5)

⇒m

2

−3m−4=0

⇒(m−4)(m+1)=0

∴m=4 or m=−1

Answer 2

Answer:

value of m for  the quadratic equation (m-5)x2 + 2(m-5)x + 2 = 0 have equal (2) roots is 5 and 9.

Step-by-step explanation:

The given equation is (m-5)x2 + 2(m-5)x + 2 = 0

For the equation to have real and equal roots, the discriminant should be 0.

discriminant D = $2(m-5)^{2}$ - 4 (m-5)×2 =0

⇒2$m^{2}$ -20m +50 - 8m + 40 = 0

⇒ 2$m^{2}$ - 28m + 90 = 0

⇒ $m^{2}$ - 14m + 45 = 0

⇒ m(m -5) - 9(m-5)=0

⇒ (m-5)(m-9) = 0

Hence , m=5 and m=9