Question

How do you find the zeros, real and imaginary, of $y = - {x^2} - 5x - 35$ using the quadratic formula?

Answer 1

Hint: First, compare the given quadratic equation to the standard quadratic equation and find the value of numbers $a$, $b$ and $c$ in the given equation. Then, substitute the values of $a$, $b$ and $c$ in the formula of discriminant and find the discriminant of the given equation. Finally, put the values of $a$, $b$ and $D$ in the roots of the quadratic equation formula and get the desired result.

Formula used: The quantity $D = {b^2} - 4ac$ is known as the discriminant of the equation $a{x^2} + bx + c = 0$ and its roots are given by
$x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$ or $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

Complete step-by-step solution:
We know that an equation of the form $a{x^2} + bx + c = 0$, $a,b,c,x \in R$, is called a Real Quadratic Equation.
The numbers $a$, $b$ and $c$ are called the coefficients of the equation.
The quantity $D = {b^2} - 4ac$ is known as the discriminant of the equation $a{x^2} + bx + c = 0$ and its roots are given by
$x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$ or $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Next, compare $ - {x^2} - 5x - 35 = 0$ quadratic equation to standard quadratic equation and find the value of numbers $a$, $b$ and $c$.
Comparing $ - {x^2} - 5x - 35 = 0$ with $a{x^2} + bx + c = 0$, we get
$a = - 1$, $b = - 5$ and $c = - 35$
Now, substitute the values of $a$, $b$ and $c$ in $D = {b^2} - 4ac$ and find the discriminant of the given equation.
$D = {\left( { - 5} \right)^2} - 4\left( { - 1} \right)\left( { - 35} \right)$
After simplifying the result, we get
$ \Rightarrow D = 25 - 140$
$ \Rightarrow D = - 115$
Which means the given equation has no real roots or has imaginary roots.
Now putting the values of $a$, $b$ and $D$ in $x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$, we get
$x = \dfrac{{ - \left( { - 5} \right) \pm \sqrt {115} i}}{{2 \times \left( { - 1} \right)}}$
It can be written as
$ \Rightarrow x = - \dfrac{{5 \pm \sqrt {115} i}}{2}$
$ \Rightarrow x = - \dfrac{5}{2} - \dfrac{{\sqrt {115} }}{2}i$ and $x = - \dfrac{5}{2} + \dfrac{{\sqrt {115} }}{2}i$
So, $x = - \dfrac{5}{2} - \dfrac{{\sqrt {115} }}{2}i$ and $x = - \dfrac{5}{2} + \dfrac{{\sqrt {115} }}{2}i$ are roots/solutions of equation $ - {x^2} - 5x - 35 = 0$.

Therefore, the imaginary zeros of $y = - {x^2} - 5x - 35$ are $x = - \dfrac{5}{2} - \dfrac{{\sqrt {115} }}{2}i$ and $x = - \dfrac{5}{2} + \dfrac{{\sqrt {115} }}{2}i$.

Note: We can check whether $x = - \dfrac{5}{2} - \dfrac{{\sqrt {115} }}{2}i$ and $x = - \dfrac{5}{2} + \dfrac{{\sqrt {115} }}{2}i$ are roots/solutions of equation $ - {x^2} - 5x - 35 = 0$ by putting the value of $x$ in given equation.
Putting $x = - \dfrac{5}{2} - \dfrac{{\sqrt {115} }}{2}i$ in LHS of equation $ - {x^2} - 5x - 35 = 0$.
\[{\text{LHS}} = - {\left( { - \dfrac{5}{2} - \dfrac{{\sqrt {115} }}{2}i} \right)^2} - 5\left( { - \dfrac{5}{2} - \dfrac{{\sqrt {115} }}{2}i} \right) - 35\]
On simplification, we get
\[ \Rightarrow {\text{LHS}} = - \dfrac{{25}}{4} + \dfrac{{115}}{4} - \dfrac{{5\sqrt {115} }}{2}i + \dfrac{{25}}{2} + \dfrac{{5\sqrt {115} }}{2}i - 35\]
\[ \Rightarrow {\text{LHS}} = 0\]
$\therefore {\text{LHS}} = {\text{RHS}}$
Thus, $x = - \dfrac{5}{2} - \dfrac{{\sqrt {115} }}{2}i$ is a solution of equation $ - {x^2} - 5x - 35 = 0$.
Putting $x = - \dfrac{5}{2} + \dfrac{{\sqrt {115} }}{2}i$ in LHS of equation $ - {x^2} - 5x - 35 = 0$.
\[{\text{LHS}} = - {\left( { - \dfrac{5}{2} + \dfrac{{\sqrt {115} }}{2}i} \right)^2} - 5\left( { - \dfrac{5}{2} + \dfrac{{\sqrt {115} }}{2}i} \right) - 35\]
On simplification, we get
\[ \Rightarrow {\text{LHS}} = - \dfrac{{25}}{4} + \dfrac{{115}}{4} + \dfrac{{5\sqrt {115} }}{2}i + \dfrac{{25}}{2} - \dfrac{{5\sqrt {115} }}{2}i - 35\]
\[ \Rightarrow {\text{LHS}} = 0\]
$\therefore {\text{LHS}} = {\text{RHS}}$
Thus, $x = - \dfrac{5}{2} + \dfrac{{\sqrt {115} }}{2}i$ is a solution of equation $ - {x^2} - 5x - 35 = 0$.
Therefore, the imaginary zeros of $y = - {x^2} - 5x - 35$ are $x = - \dfrac{5}{2} - \dfrac{{\sqrt {115} }}{2}i$ and $x = - \dfrac{5}{2} + \dfrac{{\sqrt {115} }}{2}i$.