If x = −4 Is a Root of |x 2 3; 1 x 1; 3 2 x| = 0, Then Find the Sum of the Other Two Roots

CBSE All India 2022. Previous Year Question.

Problem

If $x=-4$ is a root of $\begin{vmatrix} x & 2 & 3 \\ 1 & x & 1 \\ 3 & 2 & x \end{vmatrix}=0$, then find the sum of the other two roots.

Key insight. Expanding the determinant turns this into an ordinary cubic in $x$. Once you know one root, you don’t need to factor the whole cubic by trial — the sum of all three roots is read straight off the cubic’s coefficients, so the sum of the other two is just that total minus the known root.

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Approach

Expand the $3\times3$ determinant along the first row to get a cubic equation in $x$. Since the coefficient of $x^2$ is zero, the sum of all three roots is $0$ directly from the cubic’s structure — subtracting the known root $-4$ immediately gives the sum of the remaining two, without needing to factor the cubic at all.

Solution

Step 1 — Expand the determinant along the first row

$$D = x\begin{vmatrix}x&1\\2&x\end{vmatrix} – 2\begin{vmatrix}1&1\\3&x\end{vmatrix} + 3\begin{vmatrix}1&x\\3&2\end{vmatrix}$$

$$= x(x^2-2) – 2(x-3) + 3(2-3x)$$

Step 2 — Simplify into a cubic

$$D = x^3-2x-2x+6+6-9x = x^3-13x+12$$

Step 3 — Verify x = −4 is indeed a root

$$(-4)^3-13(-4)+12 = -64+52+12 = 0 \checkmark$$

(For contrast, $x=4$ gives $64-52+12=24\neq0$ — confirming the description’s “$x=-4$” over the title’s typo.)

Step 4 — Use the sum of all three roots

For $x^3+0\cdot x^2-13x+12=0$, the sum of all three roots is $-\dfrac{\text{coefficient of }x^2}{\text{coefficient of }x^3}=-\dfrac{0}{1}=0$.

Step 5 — Subtract the known root

$$\text{Sum of other two roots} = 0-(-4) = 4$$

Step 6 — Verify by factoring (optional check)

$$x^3-13x+12=(x+4)(x^2-4x+3)=(x+4)(x-1)(x-3)$$

Roots: $-4, 1, 3$. Sum of the other two: $1+3=4$ ✓ — matches Step 5.

Answer

Common mistakes

• Trusting a title over the actual mathematics. A video or textbook title can carry a typo (here, a missing minus sign) — checking the claimed root by direct substitution takes seconds and catches this before it derails the whole solution.
• Factoring the cubic by trial division when it isn’t necessary. Since the coefficient of $x^2$ is zero, the sum of all roots is immediately $0$ — no factoring needed to answer “sum of the other two roots” specifically.

Practise next

• If $x=-3$ is a root of $\begin{vmatrix} x & 1 & 2 \ 2 & x & 1 \ 1 & 2 & x \end{vmatrix}=0$, find the sum of the other two roots, using the same expand-then-use-the-root-sum approach.