Question

find zeros : x$x^{2} +2\sqrt{2}-6$

Answer 1

Correct Question :-

• Find zeros of quadratic polynomial $\sf{x ^2 + 2\sqrt{ 2x} - 6}.$

Given :

• Quadratic polynomial :$\sf{x ^2 + 2\sqrt{ 2x} - 6}.$

To Find :

• The zeros .

Solution :-

$: \implies \sf \: x ^{2} + 2 \sqrt{2} x - 6 = 0$

$: \implies \sf \: x ^{2} + 3 \sqrt{2} x - \sqrt{2x} - 6 = 0$

$: \implies \sf x \bigg(x + 3 \sqrt{2} \bigg) - \sqrt{2} \bigg(x + 3 \sqrt{2} \bigg) = 0$

$: \implies \sf \bigg(x - \sqrt{2} \bigg) \bigg(x + 3 \sqrt{2} \bigg) = 0$

$: \implies \sf x - \sqrt{2} = 0 \: or \: x + 3 \sqrt{2} = 0$

$: \implies \sf x = 0 + \sqrt{2} \: or \: x = 0 - 3 \sqrt{2}$

$: \implies \underline{\boxed{ \sf x = \sqrt{2} \: , \: x \: = - 3 \sqrt{2} }}$

$\therefore\underline \pink{ \sf{ Zeros \: are \: \sqrt{2} \: and \: - 3 \sqrt{2} .}}$

Answer 2

☃️Correct Question :-

Find the zeroes of polynomial :-

$\large\sf{ {x}^{2} + 2 \sqrt{2} x - 6}x$

☃️Solution :-

As per Question,

We are given with an expression and we have to find its zeroes.

So, Solution :-

$\implies \: \sf{ {x}^{2} + 2 \sqrt{2} x - 6}$

$</p><p>\implies\sf{ {x}^{2} + 3 \sqrt{2} x - \sqrt{2} x - 6}$

$\implies\sf{x(x + 3 \sqrt{2} ) - \sqrt{2} (x + 3 \sqrt{2} )}$

$\implies\sf{(x - \sqrt{2}) ( x + 3 \sqrt{2} )}$

Now, Its Zeroes :-

$\begin{gathered}(1) \sf{\: x - \sqrt{2} = 0} \\ \\ \implies\boxed{\bf\red{ \: x \: = \: \sqrt{2} }}\end{gathered} </p><p></p><p>$

$\pink{\begin{gathered}(2) \sf{\: x + 3 \sqrt{2} = 0} \\ \\ \implies\boxed{\bf\pink{x = -3 \sqrt{2} }}\end{gathered}}</p><p>$