Question

When the roots of \( x^3 + \alpha x^2 + \beta x + 6 = 0 \) are increased by 1, if one of the resultant values is the least root of \( x^4 - 6x^3 + 11x^2 - 6x = 0 \), then

• A. \( \alpha - \beta + 5 = 0 \)
• B. \( \alpha + \beta + 7 = 0 \)
• C. \( 2\alpha + \beta + 7 = 0 \)
• D. \( 2\alpha + 3\beta - 1 = 0 \) Correct Answer

Hint:

To find an equation whose roots are \(k\) more than the roots of \(f(x)=0\), replace \(x\) with \(x-k\) in \(f(x)=0\). If a value \(r_0\) is a root of \(g(y)=0\), then \(g(r_0)=0\). The least root of a polynomial is the smallest real number \(x\) for which the polynomial evaluates to zero.

Answer 1

The Correct Option is A

Step 1: Find the roots of \( x^4 - 6x^3 + 11x^2 - 6x = 0 \).
Let \( Q(x) = x^4 - 6x^3 + 11x^2 - 6x \).
Factor out \(x\): \( Q(x) = x(x^3 - 6x^2 + 11x - 6) = 0 \).
So, \(x=0\) is one root.
Let \( R(x) = x^3 - 6x^2 + 11x - 6 \).
We test for integer roots that are divisors of -6 (i.
e.
, \( \pm 1, \pm 2, \pm 3, \pm 6 \)).
\( R(1) = 1^3 - 6(1)^2 + 11(1) - 6 = 1 - 6 + 11 - 6 = 0 \).
So \(x=1\) is a root.
\( R(2) = 2^3 - 6(2)^2 + 11(2) - 6 = 8 - 24 + 22 - 6 = 0 \).
So \(x=2\) is a root.
Since \(x=1\) and \(x=2\) are roots, and the sum of roots of \(R(x)\) is \( -(-6)/1 = 6 \), if the third root is \(r_3\), then \(1+2+r_3=6 \implies r_3=3\).
Alternatively, \( R(3) = 3^3 - 6(3)^2 + 11(3) - 6 = 27 - 54 + 33 - 6 = 0 \).
So \(x=3\) is a root.
The roots of \( Q(x)=0 \) are \(0, 1, 2, 3\).
The least root of \( Q(x)=0 \) is \(0\).

Step 2: Relate the roots of the original equation to the new roots.
Let the roots of \( P(x) = x^3 + \alpha x^2 + \beta x + 6 = 0 \) be \( r_1, r_2, r_3 \).
The roots are increased by 1, so the new roots are \( r_1+1, r_2+1, r_3+1 \).
One of these resultant values is the least root of \( Q(x)=0 \), which is \(0\).
So, one of the new roots is \(0\).
Let this new root be \(y\).
If \(y\) represents a new root, and \(x\) represents an old root, then \( y = x+1 \).
The equation whose roots are \(y\) (the new roots) is obtained by substituting \( x = y-1 \) into \( P(x)=0 \).
So, \( (y-1)^3 + \alpha(y-1)^2 + \beta(y-1) + 6 = 0 \).

Step 3: Use the fact that \(y=0\) is a root of the new equation.
Substitute \( y=0 \) into the transformed equation: \[ (0-1)^3 + \alpha(0-1)^2 + \beta(0-1) + 6 = 0 \] \[ (-1)^3 + \alpha(-1)^2 + \beta(-1) + 6 = 0 \] \[ -1 + \alpha(1) - \beta + 6 = 0 \] \[ -1 + \alpha - \beta + 6 = 0 \] \[ \alpha - \beta + 5 = 0 \] This matches option (1).