Two real numbers alpha and beta are such that | vedclass

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Question

(A) Given that $\alpha + \beta = 3$ and $|\alpha - \beta| = 4$.Squaring the second equation,we get $(\alpha - \beta)^2 = 16$.We know the identity $(\a

Vedclass · Accepted Answer

$4x^2 - 12x - 7 = 0$

Vedclass · Answer

(A) Given that $\alpha + \beta = 3$ and $|\alpha - \beta| = 4$.Squaring the second equation,we get $(\alpha - \beta)^2 = 16$.We know the identity $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta$.Substituting the known values: $16 = (3)^2 - 4\alpha\beta$.$16 = 9 - 4\alpha\beta$.$4\alpha\beta = 9 - 16 = -7$.So,$\alpha\beta = -\frac{7}{4}$.The quadratic equation with roots $\alpha$ and $\beta$ is given by $x^2 - (\alpha + \beta)x + \alpha\beta = 0$.Substituting the values: $x^2 - 3x - \frac{7}{4} = 0$.Multiplying the entire equation by $4$,we get $4x^2 - 12x - 7 = 0$.

Two real numbers $\alpha$ and $\beta$ are such that $\alpha + \beta = 3$ and $|\alpha - \beta| = 4$. Then $\alpha$ and $\beta$ are the roots of the quadratic equation:

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