The solution set of a quadratic equation x^{2 | vedclass

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(D) To find the solution set of the quadratic equation $x^{2}+5x-14=0$,we factorize the quadratic expression.We look for two numbers whose product is

Vedclass · Accepted Answer

$\{-7, 2\}$

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(D) To find the solution set of the quadratic equation $x^{2}+5x-14=0$,we factorize the quadratic expression.We look for two numbers whose product is $-14$ and whose sum is $5$.These numbers are $7$ and $-2$,since $7 	imes (-2) = -14$ and $7 + (-2) = 5$.Thus,we can write the equation as: $x^{2} + 7x - 2x - 14 = 0$.Grouping the terms,we get: $x(x + 7) - 2(x + 7) = 0$.Factoring out $(x + 7)$,we have: $(x + 7)(x - 2) = 0$.Setting each factor to zero,we get $x + 7 = 0$ or $x - 2 = 0$.Therefore,$x = -7$ or $x = 2$.The solution set is $\{-7, 2\}$.

The solution set of a quadratic equation $x^{2}+5x-14=0$ is $\ldots \ldots \ldots \ldots .$

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