Question

If \( -1 \) is a twice repeated root of the equation \( a(x^3 + x^2) + bx + c = 0 \), then the ratio \( a : b : c \) is:

• A. \( 1 : -1 : 1 \)
• B. \( -1 : 1 : 1 \)
• C. \( 1 : 1 : -1 \)
• D. \( 1 : 1 : 1 \)

Hint:

When dealing with repeated roots, expand the factored form and match the coefficients to solve for unknowns.

Answer 1

The Correct Option is B

Since \( -1 \) is a repeated root of the cubic equation, we can factor the polynomial as \( a(x+1)^2(x-r) = 0 \), where \( r \) is another root. Step 1: Expand the factored form. Expanding \( (x + 1)^2(x - r) \): \[ (x + 1)^2 = x^2 + 2x + 1 \] \[ (x + 1)^2(x - r) = (x^2 + 2x + 1)(x - r) = x^3 - rx^2 + 2x^2 - 2rx + x - r \] \[ = x^3 + (2 - r)x^2 + (1 - 2r)x - r \] Now multiply by \( a \): \[ a(x^3 + (2 - r)x^2 + (1 - 2r)x - r) = a x^3 + a(2 - r)x^2 + a(1 - 2r)x - ar \] Step 2: Compare with the original equation. The original equation is \( a(x^3 + x^2) + bx + c = 0 \), which expands to: \[ a x^3 + a x^2 + bx + c = 0 \] By comparing coefficients, we get the system: - \( a = a \) (from \( x^3 \) term), - \( a(2 - r) = a \), which gives \( r = 1 \), - \( a(1 - 2r) = b \), which simplifies to \( b = a \), - \( -ar = c \), which gives \( c = -a \). Thus, the ratio of \( a : b : c \) is: \[ a : b : c = -1 : 1 : 1 \]