Question

Solve the equation $3{{y}^{2}}+8y+5=0$ using the quadratic formula.

Answer 1

Hint: Recall the quadratic formula. According to quadratic formula, the roots of the equation $a{{x}^{2}}+bx+c=0$ are given by $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. Compare the equation $3{{y}^{2}}+8y+5=0$ with the general form of the quadratic equation $a{{x}^{2}}+bx+c=0$ and hence find the value of a, b and c. Use the quadratic formula to determine the roots of the equation. Verify the solutions.

Complete step-by-step answer:
We have
$3{{y}^{2}}+8y+5=0$
Comparing with the general form of the quadratic expression $a{{x}^{2}}+bx+c=0$, we get
a =3, b = 8 and c = 5.
We know that the roots of the equation $a{{x}^{2}}+bx+c=0$ are given by $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.
Hence, we have
$y=\dfrac{-8\pm \sqrt{{{8}^{2}}-4\times 5\times 3}}{2\times 3}$
Simplifying, we get
$\begin{align}
  & y=\dfrac{-8\pm \sqrt{64-60}}{6} \\
 & \Rightarrow y=\dfrac{-8\pm 2}{6} \\
\end{align}$
Taking the positive sign, we get
$y=\dfrac{-8+2}{6}=\dfrac{-6}{6}=-1$
Taking the negative sign, we get
$y=\dfrac{-8-2}{6}=\dfrac{-10}{6}=\dfrac{-5}{3}$
Hence, we have
$y=-1,-\dfrac{5}{3}$, which are the required roots of the equation.

Note: If $\alpha $ and $\beta $ are the roots of the equation $a{{x}^{2}}+bx+c=0$, then the sum of the roots of the equation is given by $\alpha +\beta =\dfrac{-b}{a}$ and the product of the roots of the equation is given by $\alpha \beta =\dfrac{c}{a}$
Now, we have
$-1+\left( \dfrac{-5}{3} \right)=\dfrac{-3-5}{3}=-\dfrac{8}{3}$
Hence, we have
$-1+\left( \dfrac{-5}{3} \right)=\dfrac{-b}{a}$
Also, we have
$\left( -1 \right)\left( \dfrac{-5}{3} \right)=\dfrac{5}{3}$
Hence, we have
$\left( -1 \right)\left( \dfrac{-5}{3} \right)=\dfrac{c}{a}$
Hence, our solution is verified to be correct.