Question

p(x)=x^2+(3-√2)x-3√2 having one of its zeros as √2 find the other zero​

Answer 1

Answer:

The other zero is -3

Step-by-step explanation:

Given ,

p(x) = x² + (3-√2)x - 3√2

Here , a = 1 , b = (3 - √2)

and c = -3√2

$let \: the \: roots \: of \: p(x) \: be \\ \alpha \: and \: \beta \\ so \: \: \alpha = \sqrt{2} \: \: and \: \beta = ?$

We know that ,

Sum of the zeroes = -b/a

→ √2 + ß = -(3-√2)/1

→√2 + ß = -3 +√2

→ ß = -3 +√2 - √2

→ ß = -3

Therefore , the other zero is -3