
Find the zeros of the polynomial ${t^2} - 15$.
Answer
506.1k+ views
Hint: Zeros of the polynomial are the roots of the polynomial. Reduce the given function by using formula $(a^{2}-b^{2}) =(a-b)\times(a+b)$ and equate it to zero and calculate value of x.
Complete step-by-step answer:
We have been given the polynomial, ${t^2} - 15$.
So, for finding the zeros put the given polynomial equals to zero.
By using $(a^{2}-b^{2}) =(a-b)\times(a+b)$ formula
Where $a = t$ and $b=\sqrt{15}$
Then we can write $(t^{2}-{\sqrt{15}}^{2}) =(t-\sqrt{15})\times(t+\sqrt{15})$
$
\Rightarrow (t-\sqrt{15})= 0 \text{ or } (t+\sqrt{15})=0 \\
\Rightarrow t=\sqrt{15} \text{ & } t=-\sqrt{15} \\
\Rightarrow t = \pm \sqrt {15} = \pm 3.87 \\
$
Therefore, the zeros of the quadratic polynomial are $ + \sqrt {15} = 3.87$ and $ - \sqrt {15} = - 3.87$.
Note: Whenever such types of questions appear, to find the zeros or the roots of the given polynomial. Always put the polynomial given in the question equals to zero and then solve it to find the zeros.
Complete step-by-step answer:
We have been given the polynomial, ${t^2} - 15$.
So, for finding the zeros put the given polynomial equals to zero.
By using $(a^{2}-b^{2}) =(a-b)\times(a+b)$ formula
Where $a = t$ and $b=\sqrt{15}$
Then we can write $(t^{2}-{\sqrt{15}}^{2}) =(t-\sqrt{15})\times(t+\sqrt{15})$
$
\Rightarrow (t-\sqrt{15})= 0 \text{ or } (t+\sqrt{15})=0 \\
\Rightarrow t=\sqrt{15} \text{ & } t=-\sqrt{15} \\
\Rightarrow t = \pm \sqrt {15} = \pm 3.87 \\
$
Therefore, the zeros of the quadratic polynomial are $ + \sqrt {15} = 3.87$ and $ - \sqrt {15} = - 3.87$.
Note: Whenever such types of questions appear, to find the zeros or the roots of the given polynomial. Always put the polynomial given in the question equals to zero and then solve it to find the zeros.
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