Question

If 5 is one root of the equation

$\begin{vmatrix} x & 3 & 7 \\ 2 & x & -2 \\ 7 & 8 & x \end{vmatrix} = 0$, then other two roots of the equation are

• A. 2 and 7
• B. -2 and 7
• C. 2 and 7
• D. -2 and 7

Answer 1

2 and -7

Answer 2

The given equation is a determinant set to zero: $\begin{vmatrix} x & 3 & 7 \\ 2 & x & -2 \\ 7 & 8 & x \end{vmatrix} = 0$

First, we expand the determinant to find the polynomial equation in $x$: $x \begin{vmatrix} x & -2 \\ 8 & x \end{vmatrix} - 3 \begin{vmatrix} 2 & -2 \\ 7 & x \end{vmatrix} + 7 \begin{vmatrix} 2 & x \\ 7 & 8 \end{vmatrix} = 0$

$x(x \cdot x - (-2) \cdot 8) - 3(2 \cdot x - (-2) \cdot 7) + 7(2 \cdot 8 - x \cdot 7) = 0$ $x(x^2 + 16) - 3(2x + 14) + 7(16 - 7x) = 0$ $x^3 + 16x - 6x - 42 + 112 - 49x = 0$

Combine like terms: $x^3 + (16 - 6 - 49)x + (-42 + 112) = 0$ $x^3 + (10 - 49)x + 70 = 0$ $x^3 - 39x + 70 = 0$

We are given that 5 is one root of this equation. Let's verify this by substituting $x=5$ into the equation: $5^3 - 39(5) + 70 = 125 - 195 + 70 = -70 + 70 = 0$ Since the result is 0, $x=5$ is indeed a root.

Since 5 is a root, $(x-5)$ is a factor of the polynomial $x^3 - 39x + 70$. We can use polynomial division or synthetic division to find the other factor (a quadratic polynomial).

Using synthetic division with root 5:

5 | 1   0   -39   70
  |     5    25  -70
  ------------------
    1   5   -14    0

The quotient is $x^2 + 5x - 14$. So, the equation can be factored as: $(x-5)(x^2 + 5x - 14) = 0$

Now, we need to find the roots of the quadratic equation $x^2 + 5x - 14 = 0$. We can factor this quadratic equation: We look for two numbers that multiply to -14 and add up to 5. These numbers are 7 and -2. So, $x^2 + 5x - 14 = (x+7)(x-2) = 0$

Setting each factor to zero gives the other two roots: $x+7 = 0 \implies x = -7$ $x-2 = 0 \implies x = 2$

Thus, the three roots of the equation are 5, 2, and -7. Given that 5 is one root, the other two roots are 2 and -7.

Upon reviewing the provided options (0 2 and 7, 0 -2 and 7, 0 2 and 7, 0 -2 and 7), it appears there is a discrepancy as none of the options list the pair (2 and -7). The closest options are (2 and 7) or (-2 and 7). However, we verified that 7 is not a root ($7^3 - 39(7) + 70 = 343 - 273 + 70 = 140 \neq 0$) and -2 is not a root ($(-2)^3 - 39(-2) + 70 = -8 + 78 + 70 = 140 \neq 0$).

Given the situation, and assuming there might be an error in the options provided in the question, the mathematically derived correct other two roots are 2 and -7. If forced to choose from the given options, this question is flawed. However, if we assume the closest option or a partial match is expected, it's ambiguous. Since the problem asks for "other two roots", a pair is expected.

If we must select from the given options, and acknowledge the presence of an error in the question's options, there is no strictly correct choice. However, in an exam scenario, if one value matches, it might be an indication. Here, 2 is a root. But -7 is not in any pair with 2.

The only way to reconcile this is if there's a typo in the question itself. But based on the question as written, the solution is unique.

Final Answer based on calculation: The other two roots are 2 and -7. Since the options provided do not match the calculated correct answer, we state the derived answer.