Question

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

Answer 1

CORRECT QUESTION :-

★ Find zeros of quadratic polynomial x² + 2√2x - 6.

GIVEN :-

• Quadratic polynomial x² + 2√2x - 6.

TP FIND :-

• The zeros .

SOLUTION :-

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

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

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

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

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

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

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

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

Answer 2

Correct Question :-

Find the zeroes of polynomial :- $\large\sf{ {x}^{2} + 2 \sqrt{2} x - 6}$.

Solution :-

As per Question,

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

So, Solution :-

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

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

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

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

Now, Its Zeroes :-

$(1) \sf{\: x - \sqrt{2} = 0} \\ \\ \implies\boxed{\bf\red{ \: x \: = \: \sqrt{2} }}$

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