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# How do you solve ${x^2} + 5x + 7 = 0$ using a quadratic formula?

Last updated date: 24th Jun 2024
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Hint: This is a two degree one variable equation, or simply a quadratic equation. It can be solved with the help of a quadratic formula that first finds the discriminant of the given quadratic equation and then puts the values directly in the roots formula of the quadratic equation to get the required solution. You will get two solutions for a quadratic equation.
For a quadratic equation $a{x^2} + bx + c = 0$, discriminant is given as follows
$D = {b^2} - 4ac$
And its roots are given as
$x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$
Using this information we can solve the given question.

Complete step by step solution:
To solve the equation ${x^2} + 5x + 7 = 0$, using quadratic formula we will first find the discriminant of the given equation
Discriminant of a quadratic equation $a{x^2} + bx + c = 0$, is calculated as follows
$D = {b^2} - 4ac$
And roots are given as
$x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$
For the equation ${x^2} + 5x + 7 = 0$ values of $a,\;b\;{\text{and}}\;c$ are equals to
$1,\;5\;{\text{and}}\;7$ respectively,
$\therefore$ The discriminant for given equation will be
$D = {5^2} - 4 \times 1 \times 7 \\ = 25 - 28 \\ = - 3 \\$
Now finding the solution, by putting all respective values in the root formula
$x = \dfrac{{ - b \pm \sqrt D }}{{2a}} \\ = \dfrac{{ - 5 \pm \sqrt { - 3} }}{{2 \times 1}} \\ = \dfrac{{ - 5 \pm \sqrt { - 3} }}{2} \\$
Here we are getting complex form, we can write $\sqrt { - 3} = i\sqrt 3$ as we know that in complex numbers value of $i$ is equals to $\sqrt { - 1}$
So the required roots will be $x = \dfrac{{ - 5 + i\sqrt 3 }}{2}\;{\text{and}}\;x = \dfrac{{ - 5 - i\sqrt 3 }}{2}$.
These are called complex roots.

Note: Discriminant of a quadratic equation gives us the information about nature of roots of the equation, if it is greater than zero or positive then roots will be real and distinct, else if it is equals to zero then roots will be real and equal and if it is less than zero or negative then roots will be imaginary (as we have seen in this question).