Find the roots of the following quadratic equ | vedclass

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Question

(C) Given equation: $3x^2 + 11x + 10 = 0$. Divide the entire equation by $3$ to make the coefficient of $x^2$ equal to $1$: $x^2 + \frac{11}{3}x + \fr

Vedclass · Accepted Answer

$-\frac{5}{3}, -2$

Vedclass · Answer

(C) Given equation: $3x^2 + 11x + 10 = 0$. Divide the entire equation by $3$ to make the coefficient of $x^2$ equal to $1$: $x^2 + \frac{11}{3}x + \frac{10}{3} = 0$. Shift the constant term to the right side: $x^2 + \frac{11}{3}x = -\frac{10}{3}$. Add the square of half the coefficient of $x$ to both sides. The coefficient of $x$ is $\frac{11}{3}$,so half of it is $\frac{11}{6}$. The square is $(\frac{11}{6})^2 = \frac{121}{36}$. $x^2 + \frac{11}{3}x + \frac{121}{36} = -\frac{10}{3} + \frac{121}{36}$. $(x + \frac{11}{6})^2 = -\frac{120}{36} + \frac{121}{36} = \frac{1}{36}$. Taking the square root on both sides: $x + \frac{11}{6} = \pm \frac{1}{6}$. Case $1$: $x + \frac{11}{6} = \frac{1}{6} \implies x = \frac{1}{6} - \frac{11}{6} = -\frac{10}{6} = -\frac{5}{3}$. Case $2$: $x + \frac{11}{6} = -\frac{1}{6} \implies x = -\frac{1}{6} - \frac{11}{6} = -\frac{12}{6} = -2$. Thus,the roots are $-\frac{5}{3}$ and $-2$.

Find the roots of the following quadratic equation by the method of completing the square: $3x^2 + 11x + 10 = 0$.

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