Question

If equation $(k-1)x^2+(k^2+1)x+6=0and2x^2+10x+12=0$ have both roots common, the find the value of k___

Answer 1

Since both the roots are common

$\underset{\left(1\right)}{\frac{k-1}{2}}$ = $\underset{\left(2\right)}{\frac{k^2+1}{10}}$ = $\underset{\left(3\right)}{\frac{6}{12}}$

$$\begin{array}{ccc}From\left(1\right)and\left(2\right)& From\left(2\right)and\left(3\right)& From\left(1\right)and\left(3\right)\\ \frac{k-1}{2}=\frac{k^2+1}{10}& \frac{k^2+1}{10}=\frac{6}{12}& \frac{k-1}{2}=\frac{6}{12}\\ 5k-5=k^2+1& k^2=5-1& k=2\\ k^2-5k+6=0& k^2=4& \\ k=2,3& k=2,-2& \end{array}$$

The value of k is = 2 {common value of k is 2}