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
\[\begin{align}
  & a=\sqrt{3}.....(3) \\
 & b=27.......(4) \\
 & c=5\sqrt{3}.....(5) \\
\end{align}\]
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 \[\sqrt{3}{{x}^{2}}+27x+5\sqrt{3}=0\].
\[\begin{align}
  & a=\sqrt{3}.......(1) \\
 & b=27.......(2) \\
 & c=5\sqrt{3}.....(3) \\
\end{align}\]
From equation (1), (2) and (3),
\[\alpha =\dfrac{-27+\sqrt{{{(27)}^{2}}-4(\sqrt{3})(5\sqrt{3})}}{2\sqrt{3}}=\dfrac{-27+\sqrt{669}}{2\sqrt{3}}\].
\[\beta =\dfrac{-27-\sqrt{{{(27)}^{2}}-4(\sqrt{3})(5\sqrt{3})}}{2\sqrt{3}}=\dfrac{-27-\sqrt{669}}{2\sqrt{3}}\].
\[\begin{align}
  & \alpha +\beta =\dfrac{-27+\sqrt{669}}{2\sqrt{3}}+\dfrac{-27-\sqrt{669}}{2\sqrt{3}}=\dfrac{-54}{2\sqrt{3}}=-9\sqrt{3} \\
 & \alpha \beta =\left( \dfrac{-27+\sqrt{669}}{2\sqrt{3}} \right)\left( \dfrac{-27-\sqrt{669}}{2\sqrt{3}} \right)=\dfrac{{{(-27)}^{2}}-669}{{{(2\sqrt{3})}^{2}}}=\dfrac{60}{12}=5 \\
\end{align}\]
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