Question

How do you find the zeros, real and imaginary, of $y = - 3{x^2} + 3x - 18$ using the quadratic formula ?

Answer 1

Hint:Zeros of a polynomial function are the values of $x$ for which $y = 0$. For finding the zeros we substitute $y = 0$ and then we solve the equation $0 = - 3{x^2} + 3x - 18$. The resulting values of $x$ will be the zeros of the given expression. The general form of a quadratic equation is given by $a{x^2} + bx + c = 0$. We can solve this equation by using quadratic formula $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$.Real zeros are zeros which can be expressed in the form of real numbers. Imaginary zeros are zeros of an equation which does not exist and they cannot be written in the form of real numbers.

Complete step by step answer:
We are given to find zeros of $y = - 3{x^2} + 3x - 18$. To find the zeros we substitute $y = 0$ and then we solve the equation $0 = - 3{x^2} + 3x - 18$. To find the value of $x$, we have to put the values of $a$, $b$ and $c$ in the quadratic formula, that is, $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$. To get the values of $a$, $b$ and $c$ from the given equation, we compare the given equation $ - 3{x^2} + 3x - 18 = 0$ with the general form of the quadratic equation.

General form of quadratic equation is written in the form of $a{x^2} + bx + c = 0$, where $a$ is the coefficient of ${x^2}$, $b$ is the coefficient of $x$ and $c$ is the constant term. The RHS is $0$. On comparing the above rearranged equation with the general form, we observe that coefficient $a$ of ${x^2}$ is $ - 3$,co-efficient $b$ of $x$ is $3$ and the constant term $c$ is$ - 18$.Thus, $a = - 3$, $b = 3$ and $c = - 18$.

Now we put the values of $a$, $b$ and $c$ in the quadratic formula to solve for the value of $x$.
\[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} \\
\Rightarrow x = \dfrac{{ - 3 \pm \sqrt {{{(3)}^2} - 4 \times ( - 3) \times 18} }}{{2 \times ( - 3)}} \\
\Rightarrow x = \dfrac{{ - 3 \pm \sqrt {9 - ( - 216)} }}{{ - 6}} \\
\Rightarrow x = \dfrac{{ - 3 \pm \sqrt {9 + 216} }}{{ - 6}} \\
\Rightarrow x = \dfrac{{ - 3 \pm \sqrt {225} }}{{ - 6}} \\ \]
Since, the square root of $225$ is $15$, we have:
\[x = \dfrac{{ - 3 \pm 15}}{{ - 6}} \\
\Rightarrow x = \dfrac{{ - 3 + 15}}{{ - 6}}or\dfrac{{ - 3 - 15}}{{ - 6}} \\
\Rightarrow x = \dfrac{{ - 12}}{{ - 6}}or\dfrac{{ - 18}}{{ - 6}} \\
\therefore x = \dfrac{{12}}{6}or\dfrac{{18}}{6}\]
Hence, the two values of $x$ are $2$ and $3$.

Therefore, the zeros of the equation $y = - 3{x^2} + 3x - 18$ are $2$ and $3$.

Note:Here, both zeros of the given equation are real zeros. However, imaginary zeros exist when the value of $D = {b^2} - 4ac$ also known as the Discriminant, is less than 0. Imaginary zeros cannot be expressed in terms of real numbers. Here, both zeros were expressed in terms of real numbers and hence, are real zeros.