Question

Find the zeros of the polynomial expression\[{{t}^{2}}-15\]. Select the correct option:
a) \[\sqrt{15}\],\[\sqrt{15}\]
b) \[\sqrt{5}\],\[\sqrt{3}\]
c) \[-\sqrt{15}\],\[\sqrt{15}\]
d) \[-\sqrt{5}\],\[-\sqrt{3}\]

Answer 1

Hint: To find the zero solve the equation \[{{t}^{2}}-15=0\], there will be two values of t, as we have the quadratic equation.

Complete step-by-step solution -
In the question, we have to find the zeros of the polynomial expression\[{{t}^{2}}-15\].
So we know that to find the zero of the polynomial expression, we have to put that polynomial expression to zero, \[{{t}^{2}}-15=0\]
Then we will solve this quadratic equation. Now, we have the equation of the form \[{{x}^{2}}=f\left( a \right)\]
So the solutions are \[x=\sqrt{f\left( a \right)},\;\;-\sqrt{f\left( a \right)}\]
Now, we have:
\[\begin{align}
  & \Rightarrow {{t}^{2}}-15=0 \\
 & \Rightarrow {{t}^{2}}=15 \\
\end{align}\]
So here\[f\left( a \right)=15\], where t is the variable as x. Hence, the solutions are
 \[\begin{align}
  & \Rightarrow x=\sqrt{f\left( a \right)},\;\;-\sqrt{f\left( a \right)} \\
 & \Rightarrow x=\sqrt{15},\;\;-\sqrt{15} \\
\end{align}\]
So the correct answer is option c.

Note: When solving the equation of the form\[{{x}^{2}}=a\], then we have to be careful that x will have two values and that will be \[x= \pm a\]. This is because we can write \[{{x}^{2}}=a\] as
\[\begin{align}
  & \Rightarrow {{x}^{2}}=a \\
 & \Rightarrow {{x}^{2}}-a=0 \\
 & \Rightarrow (x-a)(x+a)=0 \\
 & \Rightarrow x=a\,\,\text{or}\,\,x=-a \\
\end{align}\]