Search any question & find its solution
Question:
Answered & Verified by Expert
If $\alpha, \beta$ are the roots of $x^{2}+p x-q=0$ and $\gamma, \delta$ are the roots of $\mathrm{x}^{2}-\mathrm{px}+\mathrm{r}=0$ then what is $(\beta+\gamma)(\beta+\delta)$ equal to $?$
Options:
Solution:
2077 Upvotes
Verified Answer
The correct answer is:
$\mathrm{q}+\mathrm{r}$
Since, $\alpha \& \beta$ are the roots of $x^{2}+p x-q=0$, then, $\alpha+\beta=-\mathrm{p}$...(1)
and $\alpha \beta=-\mathrm{q}$...(2)
(2) put the value of $\alpha$ from $(2)$ in $(1)$
$-\frac{q}{\beta}+\beta=-p \Rightarrow-q+\beta^{2}=-p \beta$
$\Rightarrow \beta^{2}=q-p \beta$...(3)
Since $\gamma \& \delta$ are the roots of $x^{2}-p x+r=0$, then, $\gamma+\delta=\mathrm{p}, \gamma \delta=\mathrm{r}$
Now, $(\beta+\gamma)(\beta+\delta)=\beta^{2}+\beta \delta+\beta \gamma+\gamma \delta$.
$=\beta^{2}+\beta[\gamma+\delta]+\gamma \delta .=\mathrm{q}-\mathrm{p} \beta+\mathrm{p} \beta+\mathrm{r}=\mathrm{q}-\mathrm{r}$
and $\alpha \beta=-\mathrm{q}$...(2)
(2) put the value of $\alpha$ from $(2)$ in $(1)$
$-\frac{q}{\beta}+\beta=-p \Rightarrow-q+\beta^{2}=-p \beta$
$\Rightarrow \beta^{2}=q-p \beta$...(3)
Since $\gamma \& \delta$ are the roots of $x^{2}-p x+r=0$, then, $\gamma+\delta=\mathrm{p}, \gamma \delta=\mathrm{r}$
Now, $(\beta+\gamma)(\beta+\delta)=\beta^{2}+\beta \delta+\beta \gamma+\gamma \delta$.
$=\beta^{2}+\beta[\gamma+\delta]+\gamma \delta .=\mathrm{q}-\mathrm{p} \beta+\mathrm{p} \beta+\mathrm{r}=\mathrm{q}-\mathrm{r}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.