Question

Without solving the following quadratic equation, find the value of 'm' for which the given equation has real and equal roots.
$x^2+2(m-1)x+(m+5)=0$
[4 MARKS]

Answer 1

Concept: 1 Mark
Steps: 2 Marks
Answer: 1 Marks

Given: $x^2+2(m-1)x+(m+5)=0$
Comparing it with quadratic equation $a{x}^2+bx+c=0$, we get
a=1, b=2(m-1)and c=m+5
Now we know that for real and equal roots,$b^2-4ac=0$
$\Rightarrow \left[2\right(m-1){]}^2-4\times 1\times (m+5)=0\Rightarrow 4(m^2+1-2m)-4m-20=0\Rightarrow 4m^2+4-8m-4m-20=0\Rightarrow 4m^2-12m-16=0$
$\Rightarrow m^2-3m-4=0\Rightarrow m^2-4m+m-4=0\Rightarrow m(m-4)+1(m-4)=0\Rightarrow (m+1)(m-4)=0$
Either m + 1 = 0 or m - 4 = 0
$\Rightarrow m=-1$ or m = 4
Hence, the values of m are -1 and 4.