Question

Find the sum and product of the roots of the equation $\sqrt{3}{{x}^{2}}+27x+5\sqrt{3}=0$.

Answer 1

Hint: We know that the sum of roots of a quadratic equation $a{{x}^{2}}+bx+c=0$ is equal to $\dfrac{-b}{a}$. The product of roots of a quadratic equation $a{{x}^{2}}+bx+c=0$ is equal to $\dfrac{c}{a}$. We will compare $\sqrt{3}{{x}^{2}}+27x+5\sqrt{3}=0$ with $a{{x}^{2}}+bx+c=0$. From this we will find the values of a, b and c. With the values of a, b and c we will find the sum of roots and product of roots from the above formulae.

Complete step-by-step solution -
Before solving the question,
We should know that if $\alpha$ and $\beta$ are roots of the equation $a{{x}^{2}}+bx+c=0$.
The sum of roots of $a{{x}^{2}}+bx+c$ is $\alpha +\beta =\dfrac{-b}{a}......(1)$
The product of roots of $a{{x}^{2}}+bx+c$ is $\alpha \beta =\dfrac{c}{a}......(2)$
From the question, we are given an equation $\sqrt{3}{{x}^{2}}+27x+5\sqrt{3}=0$
By comparing the equation $\sqrt{3}{{x}^{2}}+27x+5\sqrt{3}=0$ with $a{{x}^{2}}+bx+c=0$, we get

& a=\sqrt{3}.....(3) \\\ & b=27.......(4) \\\ & c=5\sqrt{3}.....(5) \\\ \end{aligned}$$ Let the roots of $$\sqrt{3}{{x}^{2}}+27x+5\sqrt{3}=0$$ are $$\alpha $$ and $$\beta $$. We know that $$\alpha +\beta =\dfrac{-b}{a}$$. In the same way, the sum of roots of $$\sqrt{3}{{x}^{2}}+27x+5\sqrt{3}=0$$ is equal to $$\alpha +\beta $$. From equation (3) and equation (4), $$\alpha +\beta =\dfrac{-b}{a}=\dfrac{-27}{\sqrt{3}}=-9\sqrt{3}$$. Therefore, the sum of roots of $$\sqrt{3}{{x}^{2}}+27x+5\sqrt{3}=0$$ is equal to $$-9\sqrt{3}$$. We know that $$\alpha \beta =\dfrac{c}{a}$$. In the same way, the product of roots of $$\sqrt{3}{{x}^{2}}+27x+5\sqrt{3}=0$$ is equal to $$\alpha \beta $$. From equation (3) and equation (5), $$\alpha \beta =\dfrac{c}{a}=\dfrac{5\sqrt{3}}{\sqrt{3}}=5$$ Therefore, the product of roots of $$\sqrt{3}{{x}^{2}}+27x+5\sqrt{3}=0$$ is equal to 5. Hence, the sum of roots of $$\sqrt{3}{{x}^{2}}+27x+5\sqrt{3}=0$$ is equal to $$-9\sqrt{3}$$ and the product of roots is equal to 5. Note: We know that if the roots of $$a{{x}^{2}}+bx+c=0$$ are $$\alpha $$ and $$\beta $$ . Then $$\alpha =\dfrac{-b+\sqrt{{{b}^{2}}-4ac}}{2a}$$ and $$\beta =\dfrac{-b-\sqrt{{{b}^{2}}-4ac}}{2a}$$. Let the roots of $$\sqrt{3}{{x}^{2}}+27x+5\sqrt{3}=0$$ are $$\alpha $$ and $$\beta $$. Let us compare $$a{{x}^{2}}+bx+c=0$$ with $$\sqrt{3}{{x}^{2}}+27x+5\sqrt{3}=0$$. $$\begin{aligned} & a=\sqrt{3}.......(1) \\\ & b=27.......(2) \\\ & c=5\sqrt{3}.....(3) \\\ \end{aligned}$$ From equation (1), (2) and (3), $$\alpha =\dfrac{-27+\sqrt{{{(27)}^{2}}-4(\sqrt{3})(5\sqrt{3})}}{2\sqrt{3}}=\dfrac{-27+\sqrt{669}}{2\sqrt{3}}$$. $$\beta =\dfrac{-27-\sqrt{{{(27)}^{2}}-4(\sqrt{3})(5\sqrt{3})}}{2\sqrt{3}}=\dfrac{-27-\sqrt{669}}{2\sqrt{3}}$$. $$\begin{aligned} & \alpha +\beta =\dfrac{-27+\sqrt{669}}{2\sqrt{3}}+\dfrac{-27-\sqrt{669}}{2\sqrt{3}}=\dfrac{-54}{2\sqrt{3}}=-9\sqrt{3} \\\ & \alpha \beta =\left( \dfrac{-27+\sqrt{669}}{2\sqrt{3}} \right)\left( \dfrac{-27-\sqrt{669}}{2\sqrt{3}} \right)=\dfrac{{{(-27)}^{2}}-669}{{{(2\sqrt{3})}^{2}}}=\dfrac{60}{12}=5 \\\ \end{aligned}$$ Hence, the sum of roots of $$\sqrt{3}{{x}^{2}}+27x+5\sqrt{3}=0$$ is equal to $$-9\sqrt{3}$$. The product of roots of $$\sqrt{3}{{x}^{2}}+27x+5\sqrt{3}=0$$ is equal to 5