Question

Given are the factors of a quadratic equation $3\left( \dfrac{5}{3}-x \right)$ and $3\left( \dfrac{4}{7}-x \right)$. Find the quadratic equation.
$\begin{align}
  & a)20-23x+21{{x}^{2}} \\
 & b)40+23x+21{{x}^{2}} \\
 & c)40+47x-21{{x}^{2}} \\
 & d)20-47x+21{{x}^{2}} \\
\end{align}$

Answer 1

Hint: Now we know that if (x – a) × (x – b) is the factor of a quadratic equation a and b will be the roots of equation and the equation is given by multiplying the terms (x – a) × (x – b) = ${{x}^{2}}-ax-bx+ab$

Complete step-by-step answer:
Now we have that $3\left( \dfrac{5}{3}-x \right)$ and $3\left( \dfrac{4}{7}-x \right)$ are factors of a quadratic equation.
Now if we take $3\left( \dfrac{5}{3}-x \right)=0$ we get $x=\dfrac{5}{3}$ and similarly if $3\left( \dfrac{4}{7}-x \right)=0$ then $x=\dfrac{4}{7}$
Then we know that $x=\dfrac{5}{3}$ and $x=\dfrac{4}{7}$ must be the roots of the required quadratic equation.
Now let us find the equation whose roots are $x=\dfrac{5}{3}$ and $x=\dfrac{4}{7}$
Now we know that for $x=\dfrac{5}{3}$ and $x=\dfrac{4}{7}$ $\left( \dfrac{5}{3}-x \right)\left( \dfrac{4}{7}-x \right)=0$
Hence multiplying the brackets we get
$\begin{align}
  & \Rightarrow \dfrac{5}{3}\times \dfrac{4}{7}-\dfrac{4}{7}x-\dfrac{5}{3}x+{{x}^{2}}=0 \\
 & \Rightarrow \dfrac{20}{21}-\dfrac{4x}{7}-\dfrac{5x}{3}+{{x}^{2}}=0 \\
\end{align}$
Let us take – 1 common for middle terms.
$\Rightarrow \dfrac{20}{21}-\left( \dfrac{4x}{7}+\dfrac{5x}{3} \right)+{{x}^{2}}=0$
Now taking LCM for terms $\dfrac{4x}{7}$ and $\dfrac{5x}{3}$ we get
$\Rightarrow \dfrac{20}{21}-\left( \dfrac{\left( 4\times 3 \right)x+\left( 5\times 7 \right)x}{21} \right)+{{x}^{2}}=0$
Now simplifying the equation we get
$\begin{align}
  & \Rightarrow \dfrac{20}{21}-\left( \dfrac{\left( 12 \right)x+\left( 35 \right)x}{21} \right)+{{x}^{2}}=0 \\
 & \Rightarrow \dfrac{20}{21}-\dfrac{47x}{21}+{{x}^{2}}=0 \\
\end{align}$
Now multiplying the equation by 21 on both sides we get.
$20-47x+21{{x}^{2}}=0$
Hence now we have a quadratic equation whose roots are $x=\dfrac{5}{3}$ and $x=\dfrac{4}{7}$ .
And hence we have $\left( \dfrac{5}{3}-x \right)$ and $\left( \dfrac{4}{7}-x \right)$ are roots of the equation $20-47x+21{{x}^{2}}$ .
Hence $3\left( \dfrac{5}{3}-x \right)$ and $3\left( \dfrac{4}{7}-x \right)$ are also factors of equation $20-47x+21{{x}^{2}}$ .

So, the correct answer is “Option d”.

Note: We can also solve this expression by checking each option. We will have to substitute $x=\dfrac{5}{3}$ and $x=\dfrac{4}{7}$ in each quadratic equation. Then for the option where we get the quadratic as 0 for both values is the correct option since it suggests that $x=\dfrac{5}{3}$ and $x=\dfrac{4}{7}$ are the roots of quadratic equation. For example if we substitute $x=\dfrac{5}{3}$ in option c that is the equation $20-47x+21{{x}^{2}}=0$ we get
$\begin{align}
  & 20-47\times \dfrac{5}{3}+21\times \dfrac{25}{9} \\
 & 20-\dfrac{235}{3}+\dfrac{525}{9} \\
 & \dfrac{180-705+525}{9} \\
 & =0 \\
\end{align}$
And also for $x=\dfrac{4}{7}$ we get
$\begin{align}
  & 20-47\times \dfrac{4}{7}+21\times \dfrac{16}{49} \\
 & 20-\dfrac{188}{7}+\dfrac{336}{49} \\
 & \dfrac{980-1316+336}{49} \\
 & =0 \\
\end{align}$