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

Step 1: Name the transformation.
The new roots are $y=\frac{x+2}{x-1}$ where $x$ is a root of $ax^2+bx+c=0$. To get the equation in $y$, solve for $x$ and substitute.
Step 2: Invert the relation.
From $y=\frac{x+2}{x-1}$, cross multiply: $y(x-1)=x+2$, so $x(y-1)=y+2$, giving $x=\frac{y+2}{y-1}$.
Step 3: Substitute into the quadratic.
Put $x=\frac{y+2}{y-1}$ into $ax^2+bx+c=0$ and multiply through by $(y-1)^2$ to clear denominators: $a(y+2)^2+b(y+2)(y-1)+c(y-1)^2=0$.
Step 4: Expand each piece.
$a(y^2+4y+4)+b(y^2+y-2)+c(y^2-2y+1)=0$.
Step 5: Group by power of $y$.
Coefficient of $y^2$: $a+b+c$. Coefficient of $y$: $4a+b-2c$. Constant: $4a-2b+c$. So $(a+b+c)y^2+(4a+b-2c)y+(4a-2b+c)=0$.
Step 6: Match the option sign convention.
Multiplying by $-1$ flips the middle sign: $(a+b+c)y^2-(4a+b-2c)y+(4a-2b+c)=0$, which is option 3.
\[ \boxed{(a+b+c)x^2-(4a+b-2c)x+(4a-2b+c)=0} \]