Question

The zeroes of the polynomial f(x)=x²-2√2x-16 are
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Answer 1

Given:

To find The zeroes of the polynomial f(x)=x²-2√2x-16 -

Solution:

Put $\[f\left( x \right) = 0\]$, for finding the zeroes of the polynomial,

$\[ \Rightarrow {x^2} - 2\sqrt 2 x - 16 = 0\]$

Factorize the equation,

$\[\begin{array}{l} \Rightarrow {x^2} - 4\sqrt 2 x + 2\sqrt 2 x - 16 = 0\\ \Rightarrow x\left( {x - 4\sqrt 2 } \right) + 2\sqrt 2 \left( {x - 4\sqrt 2 } \right) = 0\\ \Rightarrow \left( {x + 2\sqrt 2 } \right)\left( {x - 4\sqrt 2 } \right) = 0\\ \Rightarrow x = 4\sqrt 2 ,-2\sqrt 2 \end{array}\]$

Hence, the zeroes of polynomial $\[f\left( x \right) = {x^2} - 2\sqrt 2 x - 16\]$ are $\[4\sqrt 2 ,\,\,{\rm{and}} - 2\sqrt 2 \]$.

Answer 2

Step-by-step explanation:

x

2

−4

2

x+2

2

x−16=0

⇒x(x−4

2

)+2

2

(x−4

2

)=0

⇒(x+2

2

)(x−4

2

)=0

⇒x=4

2

,−2

2