Find the roots of the following quadratic equ | vedclass

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Question

(A) Given equation: $3 y^{2}+7 y-20=0$Divide the entire equation by $3$ to make the coefficient of $y^{2}$ equal to $1$:$y^{2}+\frac{7}{3} y-\frac{20}

Vedclass · Accepted Answer

$-4$ and $\frac{5}{3}$

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(A) Given equation: $3 y^{2}+7 y-20=0$Divide the entire equation by $3$ to make the coefficient of $y^{2}$ equal to $1$:$y^{2}+\frac{7}{3} y-\frac{20}{3}=0$To complete the square,add and subtract the square of half the coefficient of $y$,which is $(\frac{1}{2} 	imes \frac{7}{3})^{2} = (\frac{7}{6})^{2} = \frac{49}{36}$:$y^{2}+\frac{7}{3} y + \frac{49}{36} - \frac{49}{36} - \frac{20}{3} = 0$$(y+\frac{7}{6})^{2} - (\frac{49}{36} + \frac{240}{36}) = 0$$(y+\frac{7}{6})^{2} - \frac{289}{36} = 0$$(y+\frac{7}{6})^{2} = (\frac{17}{6})^{2}$Taking the square root on both sides:$y+\frac{7}{6} = \pm \frac{17}{6}$Case $1$: $y = \frac{17}{6} - \frac{7}{6} = \frac{10}{6} = \frac{5}{3}$Case $2$: $y = -\frac{17}{6} - \frac{7}{6} = -\frac{24}{6} = -4$Thus,the roots are $-4$ and $\frac{5}{3}$.

Find the roots of the following quadratic equation by the method of completing the square: $3 y^{2}+7 y-20=0$

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