Question

Solve for y: $\sqrt{7}{{y}^{2}}-6y-13\sqrt{7}=0$
(a)$\sqrt{7},2\sqrt{7}$
(b)$3,\dfrac{2}{\sqrt{7}}$
(c)$\dfrac{13}{\sqrt{7}},-\sqrt{7}$
(d)None of these

Answer 1

Hint: We can solve this quadratic equation by splitting it into factors. The general form of any quadratic equation is $a{{y}^{2}}+by+c=0$ , where a, b and c are constants with $a\ne 0$ . We can factorize this equation by writing b as sum or difference of the factors of the product of a and c. After we factorize, we can find the roots or solve for y by equating each factor to 0.

Complete step by step solution:
The given quadratic equation is $\sqrt{7}{{y}^{2}}-6y-13\sqrt{7}=0$. First, we compare this equation to the standard quadratic equation which is $a{{y}^{2}}+by+c=0$. By comparing, we can see that,
$\begin{align}
  & a=\sqrt{7} \\
 & b=-6 \\
 & c=-13\sqrt{7} \\
\end{align}$
Now, we need to find factors of the product of a and c.
$\begin{align}
  & a\times c=\left( \sqrt{7} \right)\times \left( -13\sqrt{7} \right) \\
 & \Rightarrow a\times c=-13\times 7 \\
\end{align}$
Therefore, the factors are -13 and 7. Now, we will try to write b as the sum of difference of -13 and 7. B can be written as,
$\begin{align}
  & b=-6 \\
 & b=-13+7 \\
\end{align}$
Therefore, the given quadratic equation becomes
$\sqrt{7}{{y}^{2}}+7y-13y-13\sqrt{7}=0$
Take terms common and find the factors. We get,
$\begin{align}
  & \sqrt{7}y\left( y+\sqrt{7} \right)-13\left( y+\sqrt{7} \right)=0 \\
 & \Rightarrow \left( \sqrt{7}y-13 \right)\left( y+\sqrt{7} \right)=0 \\
\end{align}$
Finally, equate each factor to 0 to find the roots.
$\begin{align}
  & \left( \sqrt{7}y-13 \right)=0 \\
 & \Rightarrow \sqrt{7}y=13 \\
 & \Rightarrow y=\dfrac{13}{\sqrt{7}} \\
\end{align}$
$\begin{align}
  & \left( y+\sqrt{7} \right)=0 \\
 & \Rightarrow y=-\sqrt{7} \\
\end{align}$
Hence, the solution for the given quadratic equation is $y=\dfrac{13}{\sqrt{7}},-\sqrt{7}$ .

So, the correct answer is “Option C”.

Note: Another way to find the solution for a given equation is by substituting the given options in the equation, and check if it becomes 0.
For first option (a), the quadratic function value at $y=\sqrt{7}$ is,
$\sqrt{7}{{\left( \sqrt{7} \right)}^{2}}-6\sqrt{7}-13\sqrt{7}=-12\sqrt{7}$ , clearly this is not a solution of the equation so this option is incorrect.
For second option (b), the quadratic function value at $y=3$ is,
$\sqrt{7}{{\left( 3 \right)}^{2}}-6\left( 3 \right)-13\sqrt{7}=-18-4\sqrt{7}$ , clearly this is not a solution of the equation so this option is incorrect.
For third option (c), the quadratic function value at $y=\dfrac{13}{\sqrt{7}}$ is,
$\sqrt{7}{{\left( \dfrac{13}{\sqrt{7}} \right)}^{2}}-6\left( \dfrac{13}{\sqrt{7}} \right)-13\sqrt{7}=0$
At $y=-\sqrt{7}$ , it is
$\sqrt{7}{{\left( -\sqrt{7} \right)}^{2}}-6\left( -\sqrt{7} \right)-13\sqrt{7}=0$
Hence, these 2 values are the roots of the given quadratic equation. Option (c) is correct.