Question

find the values of m such that the equation (m-4)x^2+2 (m-4)x+4=0 has equal roots.

Answer 1

hope this ll be hlpful for u.

Answer 2

Answer:

The values of m are 4 or 8.

Step-by-step explanation:

Given the equation:

$(m-4)x^2+2 (m-4)x+4=0$

A quadratic equation given by $ax^2+bx+c=0$ has equal roots, if its   discriminant is zero i.e., $b^{2} -4ac=0$

Here $a=m-4, b=2 (m-4) ~\&~ c=4$

Substituting the above in the discriminant equation, we get    

$(2 (m-4))^{2} -4\times (m-4)\times 4=0$

$\implies 4 (m^{2}-2\times m \times 4+4^{2}) +(-4m+16)\times 4=0$

$\implies 4 (m^{2}-8m +16)-16m+64=0$

$\implies 4 m^{2}-32m +64-16m+64=0$

$\implies 4m^{2}-48m +128=0$

$\implies m^{2}-12m+32=0$

Factorizing the equation,

$m^{2}-4m-8m+32=0$

$\implies m(m-4)-8(m-4)=0$

$\implies (m-8)=0~\mbox{or}~(m-4)=0$

$\implies m=8~\mbox{or}~m=4$

Therefore, the values of m are 4 or 8.