If the quadratic equation $x^{2} + b x + c$ are real numbers. Find $b$ and $c$ so that the given equation has two solutions $x = \frac{- 1}{4}$ and $x = \frac{1}{2} .$

b = \frac{- 1}{4}, c = \frac{- 1}{8}

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b = \frac{- 1}{2}, c = \frac{- 1}{8}

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b = \frac{- 1}{8}, c = \frac{- 1}{4}

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None of these

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Solution

The correct option is A $b = \frac{- 1}{4}, c = \frac{- 1}{8}$
Given,

$x^{2} + b x + c$

since, given,

$x = \frac{1}{4}, - \frac{1}{2}$ are the roots of equation, we have,

$= (x + \frac{1}{4}) (x - \frac{1}{2})$

$= x x + x (- \frac{1}{2}) + \frac{1}{4} x + \frac{1}{4} (- \frac{1}{2})$

$= x^{2} - \frac{x}{4} - \frac{1}{8}$

$∴ b = - \frac{1}{4}, c = - \frac{1}{8}$

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If the quadratic equation x2+bx+c are real numbers. Find b and c so that the given equation has two solutions x=−14 and x=12.

The correct option is A b=−14,c=−18Given,x2+bx+csince, given,x=14,−12 are the roots of equation, we have,=(x+14)(x−12)=xx+x(−12)+14x+14(−12)=x2−x4−18∴b=−14,c=−18

If the quadratic equation $x^{2} + b x + c$ are real numbers. Find $b$ and $c$ so that the given equation has two solutions $x = \frac{- 1}{4}$ and $x = \frac{1}{2} .$