Question

(X-2)^2+4. Find the zeros of the polynomial.

Answer 1

Answer:

$\large \text{ x=0 or x=4 }$

Step-by-step explanation:

Given :

$\large \text{Given (x-2)^2+4 }\\\\\\\large \text{Using identity (a-b)^2=a^2+b^2-2ab }\\\\\\\\large x^2+4-4x+4\\\\\\\large x^2-4x+8$

By using sridharacharya method we get

$\large \text{ x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a} }$

Putting values here we get

$\large x^2-4x+8\\\\\\\large \text{ x=\dfrac{4\pm\sqrt{4^2-(4\times1\times8)}}{2\times1} }\\\\\\\large \text{ x=\dfrac{4\pm\sqrt{16-32}}{2} }\\\\\\\large \text{ x=\dfrac{4\pm\sqrt{-16}}{2} }\\\\\\\large \text{ x=\dfrac{4\pm\sqrt{-(4)^2}}{2} }\\\\\\\large x=\dfrac{2(2\pm-2)}{2}\\\\\\\large x=2\pm-2\\\\\\\large \text{ x=2+(-2) or x=2-(-2) }\\\\\\\large \text{ x=0 or x=4 }$

Thus we get answer.