Question

How do you solve \[{x^2} + 8x + 15 = 0\] using the quadratic formula?

Answer 1

Hint: Here, we will compare the given equation with the standard form of a quadratic equation to find the coefficients of the equation. We will use the coefficients to find the discriminant of the given equation using the formula. Then, we will substitute the obtained discriminant and the coefficients in the quadratic formula and simplify the equation to get the required solutions of the quadratic equation.

Formula Used:
The quadratic formula states that the roots of a quadratic equation \[a{x^2} + bx + c = 0\] are given by \[x = \dfrac{{ - b \pm \sqrt D }}{{2a}}\], where \[D\] is the discriminant given by the formula \[D = {b^2} - 4ac\].

Complete step-by-step solution:
First, we will rewrite the given equation \[{x^2} + 8x + 15 = 0\].
Rewriting the given equation, we get
\[1 \cdot {x^2} + 8x + 15 = 0\]
We will use the quadratic formula to find the roots of the quadratic equation.
First, let us find the value of the discriminant.
Comparing the equation \[1 \cdot {x^2} + 8x + 15 = 0\] with the standard form of a quadratic equation \[a{x^2} + bx + c = 0\], we get
\[a = 1\], \[b = 8\], and \[c = 15\]
Substituting \[a = 1\], \[b = 8\], and \[c = 15\] in the formula for discriminant \[D = {b^2} - 4ac\], we get
\[D = {8^2} - 4\left( 1 \right)\left( {15} \right)\]
Applying the exponent 2 on the base 8, we get
\[ \Rightarrow D = 64 - 4\left( 1 \right)\left( {15} \right)\]
Multiplying the terms in the expression, we get
\[ \Rightarrow D = 64 - 60\]
Subtracting 60 from 64, we get
\[ \Rightarrow D = 4\]
Now, substituting \[a = 1\], \[b = 8\], and \[D = 4\] in the quadratic formula \[x = \dfrac{{ - b \pm \sqrt D }}{{2a}}\], we get
\[ \Rightarrow x = \dfrac{{ - 8 \pm \sqrt 4 }}{{2\left( 1 \right)}}\]
Simplifying the expression, we get
\[ \Rightarrow x = \dfrac{{ - 8 \pm 2}}{2}\]
Factoring out 2 from the numerator and simplifying, we get
\[\begin{array}{l} \Rightarrow x = \dfrac{{2\left( { - 4 \pm 1} \right)}}{2}\\ \Rightarrow x = - 4 \pm 1\end{array}\]
Therefore, either \[x = - 4 - 1\] or \[x = - 4 + 1\].
Simplifying the expressions, we get
\[x = - 5, - 3\]

Therefore, the required solutions of the given quadratic equation are \[ - 5\] and \[ - 3\].

Note:
We used the term “quadratic equation” in our solution. A quadratic equation is an equation of degree 2. It is of the form \[a{x^2} + bx + c = 0\], where \[a\] is not equal to 0. We can say that if the highest degree of an equation is 2, then the equation is termed a quadratic equation. A linear equation is an equation with the highest degree of variable as 2 and also has only 1 solution. Similarly, when the highest degree of an equation is 3, then the equation is termed a cubic equation. So, we can differentiate an equation by observing the highest degree of the equation.