Question

If the root of the equation $x^2+px+q=0$ differ from the roots of the equation $x^2+qx+p=0$ by the same quantity then find the condition.

Answer 1

We have,

$x^2+px+q=0......\left(1\right)$

$x^2+qx+p=0......\left(2\right)$

Let$\alpha and\beta$ be the roots of equation (1)

then,

$\text{Sum}\text{of}\text{roots}\alpha +\beta =\frac{-p}{1}=-p$

$\text{Product}\text{of}\text{roots}\alpha \times \beta =\frac{q}{1}=q$

Then,

$differnceofroots$

${(\alpha -\beta )}^2={(\alpha +\beta )}^2-4\alpha \beta$

${(\alpha -\beta )}^2={(-p)}^2-4q$

$(\alpha -\beta )=\sqrt{p^2-4q}$

Let ${\alpha }_{1}and{\beta }_1$ be the roots of equation (2)

$\text{sum}\text{of}\text{roots}{\alpha }_1+{\beta }_1=\frac{-q}{1}=-q$

$\text{Product}\text{of}\text{roots}{\alpha }_1.{\beta }_1=\frac{p}{1}=p$

Then,

$differnceofroots$

${({\alpha }_1-{\beta }_1)}^2={({\alpha }_1+{\beta }_1)}^2-4{\alpha }_1{\beta }_1$

${({\alpha }_1-{\beta }_1)}^2={(-q)}^2-4p$

$({\alpha }_1-{\beta }_1)=\sqrt{q^2-4p}$

According to given question,

$(\alpha -\beta )=({\alpha }_1-{\beta }_1)$

$\sqrt{p^2-4q}=\sqrt{q^2-4p}$

$\Rightarrow p^2-4q=q^2-4p\left(Squaringbothside\right)$

$\Rightarrow p^2-q^2=4q-4p$

$\Rightarrow (p+q)(p-q)=4(q-p)$

$\Rightarrow p+q=-4$

$\Rightarrow p+q+4=0$

Hence, this is the answer.