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CAT 2024 Slot 1 — QA Question 12

Two Quadratic EquationsEasy

If the equations x2 + mx + 9 = 0, x2 + nx + 17 = 0 and x2 + (m + n)x + 35 = 0 have a common negative root, then the value of (2m + 3n) is

Answer & solution

Answer: 38

Solution

Hard

A common root rr satisfies all three equations. Substitute rr into each and combine to eliminate mm and nn — the constants alone pin down rr.

1

Plug in the common root rr:

(1) r2+mr+9=0(2) r2+nr+17=0(3) r2+(m+n)r+35=0\begin{aligned} &(1)\ r^2+mr+9=0\\ &(2)\ r^2+nr+17=0\\ &(3)\ r^2+(m+n)r+35=0 \end{aligned}
2

Combine the equations to isolate the (m+n)r(m+n)r term:

2r2+(m+n)r+26=0[add (1)+(2)](m+n)r=−r2−35[rearrange (3)]\begin{aligned} &2r^2+(m+n)r+26=0 \quad\text{[add (1)+(2)]}\\ &(m+n)r=-r^2-35 \quad\text{[rearrange (3)]} \end{aligned}
3

Substitute and solve for rr:

2r2+(−r2−35)+26=0[put the (3) result into step 2]⇒ r2−9=0⇒ r=±3⇒ r=−3(negative root)\begin{aligned} &2r^2+(-r^2-35)+26=0 \quad\text{[put the (3) result into step 2]}\\ &\Rightarrow\ r^2-9=0\\ &\Rightarrow\ r=\pm3\\ &\Rightarrow\ r=-3 \quad\text{(negative root)} \end{aligned}
4

Back out m,nm,n using r=−3r=-3:

9−3m+9=0[r=−3 in (1)]⇒ m=69−3n+17=0[r=−3 in (2)]⇒ n=263⇒ 2m+3n=12+26=38\begin{aligned} &9-3m+9=0 \quad\text{[$r=-3$ in (1)]}\\ &\Rightarrow\ m=6\\ &9-3n+17=0 \quad\text{[$r=-3$ in (2)]}\\ &\Rightarrow\ n=\tfrac{26}{3}\\ &\Rightarrow\ 2m+3n=12+26=38 \end{aligned}
2m+3n=382m+3n=\mathbf{38}

“Add the first two, compare with the third” instantly kills m,nm,n and leaves r2−9=0r^2-9=0; the negative-root clue then picks r=−3r=-3.

Need a hint?

Don’t solve for m,nm,n first. A shared root means one number rr works in every equation — treat rr as the unknown and combine the equations to remove m,nm,n.

CAT 2024 Slot 1 QA Q12: If 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, — Solution | TheCATExam