Question

How do you find the zeros of $y = {x^4} - 7{x^2} + 12$?

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 $0 = {x^4} - 7{x^2} + 12$. The resulting values of $x$ will be the zeros of the given expression.

Complete step-by-step solution:
We are given to find zeros of $y = {x^4} - 7{x^2} + 12$. To find the zeros we substitute $y = 0$ and then we solve the equation $0 = {x^4} - 7{x^2} + 12$.
We now have to solve the equation ${x^4} - 7{x^2} + 12 = 0$.
This equation is a bi-quadratic equation. A bi-quadratic equation is a 4-degree equation without the terms of degree 1 and 3.
To solve such an equation we first assume ${x^2} = r$.
Putting ${x^2} = r$ in the equation we get,
$
 \Rightarrow {x^4} - 7{x^2} + 12 = 0 \\
   \Rightarrow {r^2} - 7r + 12 = 0 \\
 $
The equation is now transformed into a quadratic equation which we can solve using the quadratic formula $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$.
Substituting $a = 1$, $b = - 7$ and $c = 12$, we get,
$
  \Rightarrow r = \dfrac{{ - ( - 7) \pm \sqrt {{{( - 7)}^2} - 4 \times 1 \times 12} }}{{2 \times 1}} \\
   \Rightarrow r = \dfrac{{7 \pm \sqrt {49 - 48} }}{2} \\
   \Rightarrow r = \dfrac{{7 \pm \sqrt 1 }}{2} = \dfrac{{7 \pm 1}}{2} \\
   \Rightarrow r = \dfrac{8}{2}or\dfrac{6}{2} \\
   \Rightarrow r = 4{\kern 1pt} {\kern 1pt} {\kern 1pt} or{\kern 1pt} {\kern 1pt} {\kern 1pt} 3 \\
 $
Thus we get $r = 4{\kern 1pt} {\kern 1pt} {\kern 1pt} or{\kern 1pt} {\kern 1pt} {\kern 1pt} 3$. To get the values of $x$ from $r$ we use our assumption ${x^2} = r$.
$
\Rightarrow {x^2} = r = 4{\kern 1pt} {\kern 1pt} {\kern 1pt} or{\kern 1pt} {\kern 1pt} {\kern 1pt} 3 \\
   \Rightarrow x = \pm \sqrt 4 {\kern 1pt} {\kern 1pt} {\kern 1pt} or{\kern 1pt} {\kern 1pt} {\kern 1pt} \pm \sqrt 3 \\
   \Rightarrow x = \pm 2{\kern 1pt} {\kern 1pt} {\kern 1pt} or{\kern 1pt} {\kern 1pt} {\kern 1pt} \pm \sqrt 3 \\
   \Rightarrow x = 2,{\kern 1pt} {\kern 1pt} {\kern 1pt} - 2,{\kern 1pt} {\kern 1pt} {\kern 1pt} \sqrt 3 ,{\kern 1pt} {\kern 1pt} {\kern 1pt} or{\kern 1pt} {\kern 1pt} {\kern 1pt} - \sqrt 3 \\
 $
Thus we get the values of $x$ as $2$, $ - 2$, $\sqrt 3 $, or $ - \sqrt 3 $.
Hence, these are the zeros of the given function. All the four values are the zeros as for any value of $x$ among these the given function will yield $y = 0$.

Note: To find zeros of the function we substituted $y = 0$. This we can also understand as the x-intercept of the graph of the given function, as $y = 0$ means where the function is cutting the x-axis. A bi-quadratic equation is solved by substituting ${x^2} = r$ after which we get a quadratic equation to solve. Also when we use the value of $r$ to get the value $x$ we are actually solving another quadratic equation ${x^2} = r$. Thus we can say that a bi-quadratic equation is a combination of two quadratic equations.