Question

one zeros of the following is given,find the other zeros p(x)=x2+(3-√2) x-3√2,one zero is √2​

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

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

lettherootsofp(x)be

αandβ

soα=

2

andβ=?

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