Question

Let \(a, b, c\) be non-zero number such that \(a+b+c\neq 0\) and \(4a-2b+c\neq 0\). If \(\alpha\) and \(\beta\) are the roots of quadratic equation \(ax^2+bx+c=0\), then which of the following equation has roots \(\dfrac{\alpha+2}{\alpha-1}\) and \(\dfrac{\beta+2}{\beta-1}\)?

• A. \((a+b+c)x^2-(4a-2b+c)x+(4a+b-2c)=0\)
• B. \((a+b+c)x^2+(4a+2b-c)x+(4a-2b+c)=0\)
• C. \((a+b+c)x^2-(4a+b-2c)x+(4a-2b+c)=0\)
• D. \((a+b+c)x^2+(4a-2b+c)x+(4a+b-2c)=0\)

Hint:

For root transformations of the form \( y=\frac{ax+b}{cx+d}, \) always first express \(x\) in terms of \(y\), substitute into the original equation, and clear denominators.

Answer 1

The Correct Option is C

Concept: When roots undergo a transformation \[ y=\frac{x+2}{x-1}, \] we express \(x\) in terms of \(y\) and substitute into the original equation. This gives the transformed quadratic whose roots are the required transformed roots.

Step 1: Express \(x\) in terms of \(y\).
Let \[ y=\frac{x+2}{x-1}. \] Then \[ yx-y=x+2. \] Therefore, \[ x(y-1)=y+2. \] Hence, \[ x=\frac{y+2}{y-1}. \]

Step 2: Substitute into the original quadratic.
Given \[ ax^2+bx+c=0. \] Substituting \[ x=\frac{y+2}{y-1}, \] we get \[ a\left(\frac{y+2}{y-1}\right)^2 +b\left(\frac{y+2}{y-1}\right) +c=0. \] Multiplying by \((y-1)^2\), \[ a(y+2)^2+b(y+2)(y-1)+c(y-1)^2=0. \]

Step 3: Expand and simplify.
\[ a(y^2+4y+4) +b(y^2+y-2) +c(y^2-2y+1)=0. \] Collecting coefficients, \[ (a+b+c)y^2 +(4a+b-2c)y +(4a-2b+c)=0. \] Thus \[ (a+b+c)y^2+(4a+b-2c)y+(4a-2b+c)=0. \] Multiplying throughout by \(-1\), \[ (a+b+c)y^2-(4a+b-2c)y+(4a-2b+c)=0. \] Hence the required equation is \[ \boxed{(a+b+c)x^2-(4a+b-2c)x+(4a-2b+c)=0}. \]