Question

If \( c \) and \( d \) are the roots of \( x^2 + ax + b = 0 \), then a root of the equation} \[ x^2 + (4c + a)x + (b + 2ac + 4c^2) = 0 \] is:

• A. \( d + 2c \)
• B. \( d + c \)
• C. \( d - c \)
• D. \( \mathbf{d - 2c} \)

Hint:

If roots are transformed algebraically, try back-substitution into the new equation to verify them.

Answer 1

The Correct Option is D

Let the original roots be \( c \) and \( d \). Then: \[ c + d = -a,\quad cd = b \] Let us check if \( d - 2c \) satisfies the new equation: Plug \( x = d - 2c \) into: \[ x^2 + (4c + a)x + (b + 2ac + 4c^2) \] Use known identities: \[ x^2 = (d - 2c)^2 = d^2 - 4cd + 4c^2 \] \[ (4c + a)x = (4c + a)(d - 2c) = 4cd + ad - 8c^2 - 2ac \] Add: \[ x^2 + (4c + a)x = d^2 - 4cd + 4c^2 + 4cd + ad - 8c^2 - 2ac = d^2 + ad - 4c^2 - 2ac \] Now add the constant term: \[ b + 2ac + 4c^2 = cd + 2ac + 4c^2 \] Total expression: \[ d^2 + ad - 4c^2 - 2ac + cd + 2ac + 4c^2 = d^2 + ad + cd \] Factor: \[ d(a + d + c) = d( -c + d + c) = d^2 \Rightarrow \text{So total is } d^2 + ad + cd = 0 \text{ (since from equation)} \] \[ \Rightarrow x = d - 2c \text{ is a root} \]