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Question

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


Solution

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,
$$\text{Sum of roots}=\dfrac{-b}{a}$$
and $$\text{Product of roots}=\dfrac{c}{a}$$

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

Maths

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