Question

Find the quadratic equation whose roots exceeds the roots of quadratic equation $x^2-2x+3=0$ by $2$

Answer 1

Given that,

$x^2-2x+3=0$

Let the roots be $\alpha ,\beta$

Let the roots of the other quadratic equation be $A,B$.

Given that the roots of the other quadratic equation exceeds the roots of the given equation by $2$.

$\therefore A=\alpha +2$ and $B=\beta +2$

Now, we know that,

$\text{Sum of roots}=\frac{-b}{a}$

and $\text{Product of roots}=\frac{c}{a}$

Hence, $\alpha +\beta =2$

and $\alpha \beta =3$

Now, $A+B=(\alpha +2)+(\beta +2)=\alpha +\beta +4=6$

and, $AB=(\alpha +2)(\beta +2)=\alpha \beta +2(\alpha +\beta )+4=3+4+4=11$

Hence, $A+B=6$ and $AB=11$

Now, the quadratic equation having roots $A$ and $B$ is:

$x^2-(A+B)x+AB=0$

Substituting the values of $A+B$ and $AB$ in the equation, we get:

$x^2-6x+11=0$

Hence, the required quadratic equation is $x^2-6x+11=0$.