Question

How do you solve the equation $2{{c}^{2}}-7c=-5$ ?

Answer 1

Hint: We are given a quadratic equation which can be solved by the method of factoring the equation. We shall first find the sum of the roots and the product of the roots of the equation. Further we will find numbers which will add up to the sum of coefficient of the 2-degree term, ${{c}^{2}}$and constant term and give the result of their multiplication equal to coefficient of c, that is, -7.

Complete step by step solution:
Given that, $2{{c}^{2}}-7c=-5$
We shall first transpose the constant term -5 to the left hand side of the equation.
$\Rightarrow 2{{c}^{2}}-7c+5=0$
Since, the method of factoring the quadratic equation makes our calculations simpler, therefore, we use it the most.
For any quadratic equation $a{{x}^{2}}+bx+c=0$,
the sum of the roots $=-\dfrac{b}{a}$ and the product of the roots $=\dfrac{c}{a}$.
Thus, for the equation, $2{{c}^{2}}-7c+5=0$, $a=2,$ $b=-7$ and $c=5$.
We will find numbers by hit and trial whose product is equal to $2\times \left( 5 \right)=10$ and whose sum is equal to -7.
Such two numbers are -2 and -5 as $-2+\left( -5 \right)=-7$ and $-2\times -5=10$.
Now, factoring the equation:
$\Rightarrow 2{{c}^{2}}-2c-5c+5=0$
Taking common, we get:
$\begin{align}
  & \Rightarrow 2c\left( c-1 \right)-5\left( c-1 \right)=0 \\
 & \Rightarrow \left( c-1 \right)\left( 2c-5 \right)=0 \\
\end{align}$
Hence, $c-1=0$ or $2c-5=0$
On transposing the constant terms to the other side and then dividing it with the coefficient of c, we get
$\Rightarrow c=1$ or $c=\dfrac{5}{2}$
Therefore, the roots of the given quadratic equation are $c=1,\dfrac{5}{2}$.

Note: It is very important to reverse the sign of terms which are being transposed in an equation. We could have transposed -5 and mistakenly written it as -5 only instead of writing 5. We must acknowledge that the constant term plays an important role in determining the final solution and thus, it must be transposed with proper attention.