Question

If the equations $a_1{x}^2+b_{1}x+c_1=0$ and $a_2{x}^2+b_{2}x+c_2=0$ have a common root, then the value of $(a_1{b}_2-a_2{b}_1)(b_1{c}_2-b_2{c}_1)$ is

• A. $-(a_1{c}_2-a_2{c}_1{)}^2$
• B. $(a_1{a}_2-a_2{c}_1{)}^2$
• C. $(c_1{c}_2-a_1{a}_2{)}^2$
• D. $(a_2{c}_1-a_1{c}_2{)}^2$

Answer 1

The correct option is D $(a_2{c}_1-a_1{c}_2{)}^2$

$a_1{x}^2+b_{1}x+c_1=0$ ...(i)
$a_2{x}^2+b_{2}x+c_2=0$ ...(ii)
Multiplying the first with $a_2$ and the second with $a_1$
Subtracting equation (ii), from (i), we get
$x(a_2{b}_1-a_1{b}_2)+a_2{c}_1-a_1{c}_2=0$
$x=\frac{a_1{c}_2-a_2{c}_1}{a_2{b}_1-b_2{a}_1}$ ...(iii)
is the common root.
Now ,
Multiplying the first with $b_2$ and the second with $b_1$ and then subtracting, we get
$m^2(a_1{b}_2-a_2{b}_1)+c_1{b}_2-c_2{b}_1=0$
$m^2=\frac{b_1{c}_2-b_2{c}_1}{a_1{b}_2-b_2{a}_1}$ ...(iv)
Comparing (iii) and (iv) , we get
$(\frac{a_1{c}_2-a_2{c}_1}{a_2{b}_1-b_2{a}_1}{)}^2=\frac{b_1{c}_2-b_2{c}_1}{a_1{b}_2-b_2{a}_1}$
$(a_1{c}_2-a_2{c}_1{)}^2=(a_2{b}_1-b_2{a}_1\left)\right(b_1{c}_2-b_2{c}_1)$