Question

If both $\left( {x - 2} \right)$ and $x - \dfrac{1}{2}$ are factors of $p{x^2} + 5x + r$, show that $p = r$.

Answer 1

Hint:
When we will substitute the value of roots in the given quadratic equation, they will give the value of the polynomial as 0. Hence, we will substitute the values one by one in the given expression to get a relation in $p$ and $r$.

Complete step by step solution:
The roots of the expression are $\left( {x - 2} \right)$ and $x - \dfrac{1}{2}$
We will equate each root to 0 to find the value of $x$
$
\Rightarrow x - 2 = 0 \\
\Rightarrow x = 2 \\
$
And
$
\Rightarrow x - \dfrac{1}{2} = 0 \\
\Rightarrow x = \dfrac{1}{2} \\
$
Now, we will substitute the value of $x = 2$ in the given expression,
$
   \Rightarrow p{\left( 2 \right)^2} + 5\left( 2 \right) + r = 0 \\
   \Rightarrow 4p + 10 + r = 0 \\
$
$ \Rightarrow 4p + r = - 10$ eqn. (1)
Now, put $x = \dfrac{1}{2}$ in the given polynomial.
$
   \Rightarrow p{\left( {\dfrac{1}{2}} \right)^2} + 5\left( {\dfrac{1}{2}} \right) + r = 0 \\
   \Rightarrow \dfrac{p}{4} + \dfrac{5}{2} + r = 0 \\
   \Rightarrow p + 10 - 4r = 0 \\
$
$ \Rightarrow p - 4r = - 10$ eqn. (2)
Here, we can see that the RHS of both the equations are the same.
Therefore, we can equate equations(1) and (2).
$
   \Rightarrow 4p + r = p - 4r \\
   \Rightarrow 3p = 3r \\
   \Rightarrow p = r \\
$

Note:
We can also multiply the given roots and then compare it with the given expression to get the required relation. Also, if we have a quadratic equation, say, $a{x^2} + bx + c = 0$, then the sum of roots is $ - b$ and the product of roots is $c$.