# What is the value of c such that: x^2 + 14x + c, is a perfect-square trinomial?

Aug 9, 2016

Consider the quadratic equation ${x}^{2} + 4 x + 4 = 0$, which, on the left side, is also a perfect square trinomial. Factoring to solve:

$\implies \left(x + 2\right) \left(x + 2\right) = 0$

$\implies x = - 2 \mathmr{and} - 2$

Two identical solutions! Recall that the solutions of a quadratic equation are the x intercepts on the corresponding quadratic function.

So, the solutions to the equation ${x}^{2} + 5 x + 6 = 0$, for example, will be the x intercepts on the graph of $y = {x}^{2} + 5 x + 6$.

Similarly, the solutions to the equation ${x}^{2} + 4 x + 4 = 0$ will be the x intercepts on the graph of $y = {x}^{2} + 4 x + 4$.

Since there is really only one solution to ${x}^{2} + 4 x + 4 = 0$, the vertex of the function $y = {x}^{2} + 4 x + 4$ lies on the x axis.

Now, think of the discriminant of a quadratic equation. If you don't have previous experience with it, don't fret.

We use the discriminant, ${b}^{2} - 4 a c$, to verify how many solutions, and the solution type, a quadratic equation of the form $a {x}^{2} + b x + c = 0$ may have without solving the equation.

When the discriminant equals less than $0$, the equation will have no solution. When the discriminant equals exactly zero, the equation will have exactly one solution. When the discriminant equals any number more than zero, there will be exactly two solutions. If the number in question that you get as a result is a perfect square in the latter case, the equation will have two rational solutions. If not, it will have two irrational solutions.

I've already shown that when you have a perfect square trinomial, you will have two identical solutions, which is equal to one solution. Hence, we can set the discriminant to $0$ and solve for $c$.

Where a = 1, b = 14 and c = ?:

${b}^{2} - 4 a c = 0$

${14}^{2} - 4 \times 1 \times c = 0$

$196 - 4 c = 0$

$4 c = 196$

$c = 49$

Thus, the perfect square trinomial with $a = 1 \mathmr{and} b = 14$ is ${x}^{2} + 14 x + 49$. We can verify this by factoring.

${x}^{2} + 14 x + 49 = \left(x + 7\right) \left(x + 7\right) = {\left(x + 7\right)}^{2}$

Practice exercises:

1. Using the discriminant, determine the values of $a , b , \mathmr{and} c$ that render the trinomials perfect squares.

a) $a {x}^{2} - 12 x + 4$

b) $25 {x}^{2} + b x + 64$

c) $49 {x}^{2} + 14 x + c$

Hopefully this helps, and good luck!