Question

How do you find the zeros, real and imaginary, of $y = - {x^2} - 3x + 11$ using the quadratic formula?

Answer 1

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 $- {x^2} - 3x + 11 = 0$ quadratic equation to standard quadratic equation and find the value of numbers $a$, $b$ and $c$.
Comparing $- {x^2} - 3x + 11 = 0$ with $a{x^2} + bx + c = 0$, we get
$a = - 1$, $b = - 3$ and $c = 11$
Now, substitute the values of $a$, $b$ and $c$ in $D = {b^2} - 4ac$ and find the discriminant of the given equation.
$D = {\left( { - 3} \right)^2} - 4\left( { - 1} \right)\left( {11} \right)$
After simplifying the result, we get
$\Rightarrow D = 9 + 44$
$\Rightarrow D = 53$
Which means the given equation has real roots.
Now putting the values of $a$, $b$ and $D$ in $x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$, we get
$x = \dfrac{{ - \left( { - 3} \right) \pm \sqrt {53} }}{{2 \times \left( { - 1} \right)}}$
It can be written as
$\Rightarrow x = - \dfrac{{3 \pm \sqrt {53} }}{2}$
$\Rightarrow x = - \dfrac{3}{2} - \dfrac{{\sqrt {53} }}{2}$ and $x = - \dfrac{3}{2} + \dfrac{{\sqrt {53} }}{2}$
So, $x = - \dfrac{3}{2} - \dfrac{{\sqrt {53} }}{2}$ and $x = - \dfrac{3}{2} + \dfrac{{\sqrt {53} }}{2}$ are roots/solutions of equation $- {x^2} - 3x + 11 = 0$.
Therefore, the real zeros of $y = - {x^2} - 3x + 11$ are $x = - \dfrac{{3 \pm \sqrt {53} }}{2}$.

Note: We can check whether $x = - \dfrac{{3 \pm \sqrt {53} }}{2}$ are roots/solutions of equation $- {x^2} - 5x - 35 = 0$ by putting the value of $x$ in given equation.
Putting $x = - \dfrac{3}{2} - \dfrac{{\sqrt {53} }}{2}$ in LHS of equation $- {x^2} - 3x + 11 = 0$.
${\text{LHS}} = - {\left( { - \dfrac{3}{2} - \dfrac{{\sqrt {53} }}{2}} \right)^2} - 3\left( { - \dfrac{3}{2} - \dfrac{{\sqrt {53} }}{2}} \right) + 11$
On simplification, we get
$\Rightarrow {\text{LHS}} = - \dfrac{9}{4} - \dfrac{{53}}{4} - \dfrac{{3\sqrt {53} }}{2} + \dfrac{9}{2} + \dfrac{{3\sqrt {53} }}{2} + 11$
$\Rightarrow {\text{LHS}} = 0$
$\therefore {\text{LHS}} = {\text{RHS}}$
Thus, $x = - \dfrac{3}{2} - \dfrac{{\sqrt {53} }}{2}$ is a solution of equation $- {x^2} - 3x + 11 = 0$.
Putting $x = - \dfrac{3}{2} + \dfrac{{\sqrt {53} }}{2}$ in LHS of equation $- {x^2} - 3x + 11 = 0$.
${\text{LHS}} = - {\left( { - \dfrac{3}{2} + \dfrac{{\sqrt {53} }}{2}} \right)^2} - 3\left( { - \dfrac{3}{2} + \dfrac{{\sqrt {53} }}{2}} \right) + 11$
On simplification, we get
$\Rightarrow {\text{LHS}} = - \dfrac{9}{4} - \dfrac{{53}}{4} + \dfrac{{3\sqrt {53} }}{2} + \dfrac{9}{2} - \dfrac{{3\sqrt {53} }}{2} + 11$
$\Rightarrow {\text{LHS}} = 0$
$\therefore {\text{LHS}} = {\text{RHS}}$
Thus, $x = - \dfrac{3}{2} + \dfrac{{\sqrt {53} }}{2}$ is a solution of equation $- {x^2} - 3x + 11 = 0$.
Final solution: Therefore, the real zeros of $y = - {x^2} - 3x + 11$ are $x = - \dfrac{{3 \pm \sqrt {53} }}{2}$.