Question

How do you solve the equation ${x^2} + 8 = - 6x$ by graphing?

Answer 1

Hint: In order to solve the given quadratic equation, remember the graph of every quadratic equation is a parabola. If the coefficient of ${x^2}$ is positive then it is an upward open parabola otherwise it is a downward open parabola. The x-intercepts are the root or solution of the equation and y-intercept can be obtained by putting $x = 0$ in the equation.

Complete step by step solution:
We are given a quadratic equation as ${x^2} + 8 = - 6x$.
We can write the given equation as: $y = {x^2} + 8 + 6x$
As per the question, we have to find the important parts of the equation which will give us a better picture of the graph.
As we know, the graph of every quadratic equation is always a parabola. Parabola is basically a vague ’U’ Shaped graph.
Since the coefficient of variable $x$ is positive, we can conclude that the parabola will be an upward opening parabola.
Now, to check how many times this parabola will intersect with the x-axis, we have to find the determinant $D$ of the equation.
Let’s first compare the given equation $y = {x^2} + 6x + 8$ with the standard quadratic equation $a{x^2} + bx + c$ to get the values of $a,b,c$, we get
$
 a = 1 \\
 b = 6 \\
 c = 8 \\
 $
Determinant $D$ of quadratic equation is given as $D = {b^2} - 4ac$
Putting the values of $a,b,c$ we get the determinant as
\[
 D = {\left( 6 \right)^2} - 4\left( 1 \right)\left( 8 \right) \\
 D = 36 - 32 \\
 D = 4 \\
 \]
Since, we got $D > 0$, which means there are two distinct real roots or in other words the equation has two x-intercepts.
The Root or intercepts of x-axis are $x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$
$
 \Rightarrow {x_1} = \dfrac{{ - 6 + \sqrt 4 }}{{2\left( 1 \right)}} \\
 {x_1} = \dfrac{{ - 6 + 2}}{2} \\
 {x_1} = \dfrac{{ - 4}}{2} = - 2 \\
 \Rightarrow {x_2} = \dfrac{{ - 6 - \sqrt 4 }}{{2\left( 1 \right)}} \\
 {x_2} = \dfrac{{ - 6 - 2}}{2} = \dfrac{{ - 8}}{2} = - 4 \\
 {x_2} = - 4 \\
 $
$x = - 2\,and\,x = - 4$ are the intercepts of x-axis which we can plot.
For the y-intercept put $x = 0$ in the equation $y = {x^2} + 6x + 8$, we get
$
 y = {x^2} + 6x + 8 \\
 y = {\left( 0 \right)^2} + 0 + 8 \\
 y = 8 \\
 $
$y = 8$ this is our y-intercept.
Now using the x-intercepts as $x = - 2and\,x = - 4$and y-intercept as $y = 8$, we can graph our upward opening parabola as

Additional Information:
Quadratic Equation: A quadratic equation is an equation which can be represented in the form of $a{x^2} + bx + c$ where $x$ is the unknown variable and a,b,c are the numbers known where $a \ne 0$. If $a = 0$ then the equation will become a linear equation and will no longer be quadratic .
The degree of the quadratic equation is of the order 2.
Every Quadratic equation has 2 roots.
Discriminant: $D = {b^2} - 4ac$
Using Discriminant, we can find out the nature of the roots
> If D is equal to zero, then both of the roots will be the same and real.
> If D is a positive number then, both of the roots are real solutions.
> If D is a negative number, then the root are the pair of complex solutions

Note:
1. If the coefficient of ${x^2}$ is negative then the parabola will be a downward opening parabola.
2. To obtain the root of the equation, you can also use the identity ${A^2} - {B^2} = \left( {A - B} \right)\left( {A + B} \right)$ in the equation and put $y = 0$.
3. Don’t jump steps in such types of problems, as it may increase the chance of error in the solution.