Question

If \(\alpha,\beta\) are the roots of \(ax^2+bx+c=0\) and \(\alpha+h,\beta+h\) are the roots of \(px^2+qx+r=0\), then \(h=\)

• A. \(\left[\frac{b}{a}-\frac{q}{p}\right]\)
• B. \(\frac{1}{2}\left[\frac{b}{a}-\frac{q}{p}\right]\)
• C. \(-\frac{1}{2}\left[\frac{a}{b}-\frac{p}{q}\right]\)
• D. \(\frac{1}{2}\left[\frac{a}{b}-\frac{p}{q}\right]\)

Hint:

When roots are shifted by \(h\), compare the sum of roots: \[ (\alpha+h)+(\beta+h)=\alpha+\beta+2h \]

Answer 1

The Correct Option is B

Concept:
This question is based on the relation between roots and coefficients of a quadratic equation. For a quadratic equation: \[ Ax^2+Bx+C=0 \] If its roots are \(r_1\) and \(r_2\), then: \[ r_1+r_2=-\frac{B}{A} \] Here, the roots of the second equation are obtained by adding \(h\) to each root of the first equation.

Step 1: Use the sum of roots of the first quadratic equation.
Given: \[ ax^2+bx+c=0 \] Its roots are: \[ \alpha,\beta \] Therefore: \[ \alpha+\beta=-\frac{b}{a} \]

Step 2: Use the sum of roots of the second quadratic equation.
Given: \[ px^2+qx+r=0 \] Its roots are: \[ \alpha+h,\quad \beta+h \] So, the sum of its roots is: \[ (\alpha+h)+(\beta+h) \] \[ =\alpha+\beta+2h \] Also, from the equation \(px^2+qx+r=0\), the sum of roots is: \[ -\frac{q}{p} \] Therefore: \[ \alpha+\beta+2h=-\frac{q}{p} \]

Step 3: Substitute the value of \(\alpha+\beta\).
From
Step 1: \[ \alpha+\beta=-\frac{b}{a} \] Substitute this into: \[ \alpha+\beta+2h=-\frac{q}{p} \] So: \[ -\frac{b}{a}+2h=-\frac{q}{p} \]

Step 4: Solve for \(h\).
Move \(-\frac{b}{a}\) to the right-hand side: \[ 2h=-\frac{q}{p}+\frac{b}{a} \] Rearrange: \[ 2h=\frac{b}{a}-\frac{q}{p} \] Now divide both sides by \(2\): \[ h=\frac{1}{2}\left[\frac{b}{a}-\frac{q}{p}\right] \] Hence, the correct answer is: \[ \boxed{(B)\ \frac{1}{2}\left[\frac{b}{a}-\frac{q}{p}\right]} \]