Question

How do you find the value of c that makes ${{x}^{2}}+5x+c$ into a perfect square?

Answer 1

In this question, we have to find the value of c. As it is given that the equation is of a quadratic form. Thus, we apply completing the square method, such that the equation will become a perfect square. First, we find the value of c, which is the square of the half of the coefficient of x, then after the new equation formed, we write the equation in the form of ${{x}^{2}}+bx=c$ , and try to make the LHS a perfect square, which is our required answer.

Complete step by step answer:
According to the question, the equation is given as ${{x}^{2}}+5x+c$ , where we have to find the value of c to make it a perfect square.
Equation: ${{x}^{2}}+5x+c$
The quadratic equation is in the form of: ${{a}^{2}}+bx+c=0$ ,
Here, a=1, b=5 and c=c
So, we will apply completing the square method,
Now, c is that constant, which is the square of half of the coefficient of the linear x , such that
$\begin{aligned} & c={{\left( \dfrac{b}{2} \right)}^{2}} \\\ & \Rightarrow c={{\left( \dfrac{5}{2} \right)}^{2}} \\\ \end{aligned}$
Therefore, we get

\Rightarrow c=\dfrac{25}{4} \\\ {} \\\ \end{array}$$ So, the new equation becomes as, ${{x}^{2}}+5x+c={{x}^{2}}+5x+\dfrac{25}{4}$ Now, we write the above equation in the form of ${{x}^{2}}+bx=c$ , we get $\begin{aligned} & {{x}^{2}}+5x+\dfrac{25}{4} \\\ & \Rightarrow {{x}^{2}}+5x=-\dfrac{25}{4} \\\ \end{aligned}$ Now, add the square of half of the coefficient of x on both sides of the equation, we get $\Rightarrow {{x}^{2}}+5x+{{\left( \dfrac{5}{2} \right)}^{2}}=-\dfrac{25}{4}+{{\left( \dfrac{5}{2} \right)}^{2}}$ On further simplification, we get $\begin{aligned} & \Rightarrow {{x}^{2}}+5x+\left( \dfrac{25}{4} \right)=-\dfrac{25}{4}+\left( \dfrac{25}{4} \right) \\\ & \Rightarrow {{x}^{2}}+5x+\left( \dfrac{25}{4} \right)=0 \\\ \end{aligned}$ Therefore, we get ${{\left( x+\left( \dfrac{5}{2} \right) \right)}^{2}}=0=RHS$ Thus, we see that the value of c $=\dfrac{25}{4}$ makes the equation a perfect square root. Therefore, the value of c is $\dfrac{25}{4}$ for the equation ${{x}^{2}}+5x+c$, and its perfect square is ${{\left( x+\dfrac{5}{2} \right)}^{2}}$ **Note:** Always do calculations carefully to avoid mistakes and confusion. You can also find the perfect square by using the discriminant formula $\sqrt{{{b}^{2}}-4ac}$ , where you simply put the value of a, b, and c in the formula to get the square root of the equation.