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CAT 2024 Slot 1 QA Question & Solution

AlgebraHard

Question

lf the equations $x^{2}+mx+9=0, x^{2}+nx+17=0$ and $x^{2}+(m+n)x+35=0$ have a common negative root, then the value of $(2m+3n)$ is

Solution

When given more than one equations, stating the fact that there is a common root, 
We need to equate the two equations to get discernible values for $x$

Here, we are given three equations with the values of $m$, $n$

$x^2+mx+9=x^2+\left(m+n\right)x+35$
$mx+9=mx+nx+35$
$nx=-26$

Similarly, we can do it for the other equation as well, 
$x^2+nx+17=x^2+\left(m+n\right)x+35$
$mx=-18$

Substituting the value of either $mx$ or $nx$ in the original equations, we get 
$x^2-18+9=0$
$x^2=9$
$x=\pm\ 3$
Since we are given that the root is negative, $x=-3$

$n=-\dfrac{26}{-3}$
$m=-\dfrac{18}{-3}$

$3n=26$
$2m=12$

$2m+3n=38$