Question

The equation having the multiple root of the equation $x^4 + 4x^3 - 16x - 16 = 0$ as its root is

• A. $x^2 + 2x - 3 = 0$
• B. $x^2 - 3x + 2 = 0$
• C. $x^2 + x - 2 = 0$
• D. $x^2 - 4x + 3 = 0$

Hint:

A key property in the theory of equations is that a multiple root of $f(x)=0$ is also a root of $f'(x)=0$. This reduces the problem of finding a multiple root of a high-degree polynomial to finding a common root between the polynomial and its derivative, which is of a lower degree and often easier to solve.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
If a polynomial equation $f(x)=0$ has a multiple root (a root with multiplicity greater than 1), then that root is also a root of the derivative equation, $f'(x)=0$. Therefore, the multiple root must be a common root of $f(x)=0$ and $f'(x)=0$.
Step 2: Key Formula or Approach
1. Let $f(x) = x^4 + 4x^3 - 16x - 16$. 2. Find the derivative, $f'(x)$. 3. Solve the equation $f'(x)=0$ to find the potential multiple roots. 4. Substitute these potential roots back into the original equation $f(x)=0$ to identify the actual multiple root. 5. Check which of the given quadratic equations has this multiple root as one of its roots.
Step 3: Detailed Explanation
1. Given $f(x) = x^4 + 4x^3 - 16x - 16$.
2. Find the derivative: \[ f'(x) = 4x^3 + 12x^2 - 16 \]
3. Solve $f'(x) = 0$: \[ 4x^3 + 12x^2 - 16 = 0 \] Divide by 4 to simplify: \[ x^3 + 3x^2 - 4 = 0 \] Let's find the roots of this cubic equation by testing integer divisors of -4 (i.e., $\pm 1, \pm 2, \pm 4$). Let $g(x) = x^3 + 3x^2 - 4$.
$g(1) = 1 + 3 - 4 = 0$. So, $x=1$ is a root.
$g(-2) = (-2)^3 + 3(-2)^2 - 4 = -8 + 3(4) - 4 = -8 + 12 - 4 = 0$. So, $x=-2$ is a root.
Since it's a cubic, there's one more root. As the sum of coefficients is 0, one root is 1. We found another root is -2. The sum of roots is -3. So $1 + (-2) + r_3 = -3 \implies -1 + r_3 = -3 \implies r_3 = -2$.
So the roots of $f'(x)=0$ are $1, -2, -2$.

4. Identify the multiple root of $f(x)$:
The possible multiple roots of $f(x)=0$ are the roots of $f'(x)=0$, which are $x=1$ and $x=-2$. We test them in $f(x)=0$.
Test $x=1$:
$f(1) = (1)^4 + 4(1)^3 - 16(1) - 16 = 1 + 4 - 16 - 16 = -27 \neq 0$.
So, $x=1$ is not the multiple root of $f(x)$. Test $x=-2$:
$f(-2) = (-2)^4 + 4(-2)^3 - 16(-2) - 16 = 16 + 4(-8) + 32 - 16 = 16 - 32 + 32 - 16 = 0$.
Since $f(-2)=0$ and $f'(-2)=0$, $x=-2$ is a multiple root of $f(x)=0$.
5. Find the correct equation:
We need to find which of the given options has $x=-2$ as a root.
(A) $x^2 + 2x - 3 = 0 \implies (-2)^2 + 2(-2) - 3 = 4 - 4 - 3 = -3 \neq 0$.
(B) $x^2 - 3x + 2 = 0 \implies (-2)^2 - 3(-2) + 2 = 4 + 6 + 2 = 12 \neq 0$.
(C) $x^2 + x - 2 = 0 \implies (-2)^2 + (-2) - 2 = 4 - 2 - 2 = 0$. This is correct.
(D) $x^2 - 4x + 3 = 0 \implies (-2)^2 - 4(-2) + 3 = 4 + 8 + 3 = 15 \neq 0$.
Step 4: Final Answer
The multiple root of the given equation is $x=-2$. The equation from the options that has $x=-2$ as a root is $x^2+x-2=0$.