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Ifthe roots of a quadratic equation are $\mathrm{m}+\mathrm{n}$ and $\mathrm{m}-\mathrm{n}$, then the quadratic equation will be
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The correct answer is:
$x^{2}-2 m x+m^{2}-n^{2}=0$
Let $\alpha=\mathrm{m}+\mathrm{n}, \beta=\mathrm{m}-$
Now, $\alpha+\beta=\mathrm{m}+\mathrm{n}+\mathrm{m}-\mathrm{n}=2 \mathrm{~m}$
$\alpha \times \beta=(m+n)(m-n)=m^{2}-n^{2}$
Now, Required Quadratic Equation will be, $\mathrm{x}^{2}-(\alpha+\beta) \mathrm{x}+\alpha \beta=0$
$x^{2}-2 m x+m^{2}-n^{2}=0$
Now, $\alpha+\beta=\mathrm{m}+\mathrm{n}+\mathrm{m}-\mathrm{n}=2 \mathrm{~m}$
$\alpha \times \beta=(m+n)(m-n)=m^{2}-n^{2}$
Now, Required Quadratic Equation will be, $\mathrm{x}^{2}-(\alpha+\beta) \mathrm{x}+\alpha \beta=0$
$x^{2}-2 m x+m^{2}-n^{2}=0$
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