Question

If the equations $x^2+bx-a=0$ and $x^2-ax+b=0$ have a common root, then

• A. $a+b=0$
• B. $a=b$
• C. $a-b=1$
• D. $a+b=1$

Answer 1

Answer: (A), (C)

$a+b=0$
$a-b=1$

Only one root is common: let $\alpha$ be the common root $\frac{{\alpha }^2}{b_1{c}_2-b_2{c}_1}=\frac{\alpha }{a_2{c}_1-a_1{c}_2}=\frac{1}{a_1{b}_2-a_2{b}_1}$

Above formula is used to find common root between

$a_1{x}^2+b_{1}x+c_1=0$

$a_2{x}^2+b_{2}x+c_2=0$

Now comparing given equations with above mentioned standard equation

we get

$a_1=1,b_1=b,c_1=-a$ and $a_2=1,b_2=-a,c_2=b$

Now Putting above values in the given $Formula$ we get

$\frac{{\alpha }^2}{b^2-a^2}=\frac{\alpha }{-a-b}=\frac{1}{-a-b}$

Now ${\alpha }^2=\frac{b^2-a^2}{-a-b}$ ......................(1)

and $\alpha =1$ ......................(2)

Putting Value of $\alpha$ in (1) we get ,

$a^2-b^2$ = $a+b$

$\Rightarrow$ $(a-b)(a+b)=(a+b)$

$\Rightarrow$ $(a-b)(a+b)$ $-(a+b)=0$ ......................(3)

$\Rightarrow$$(a+b)\left[\right(a-b)-1]=0$

$\Rightarrow$ $(a+b)=0$ or $(a-b)=1$

Therefore answer is $(A,C)$