Question

4x²+5√2x-3 ka shunyak nikalen​

Answer 1

Step-by-step explanation:

$4 {x}^{2} + 5 \sqrt{2} x - 3 \\ 4 {x}^{2} + x(6 \sqrt{2} - \sqrt{2} ) - 3 \\ 4 {x}^{2} + 6 \sqrt{2} x + \sqrt{2} x - 3 \\ 2 \sqrt{2} x ( \sqrt{2}x - 3) + 1( \sqrt{2} x - 3) \\ (\sqrt{2} x - 3)(2 \sqrt{2}x + 1) = 0 \\ \sqrt{2} x - 3 = 0 \\ x = \frac{3}{ \sqrt{2} } \\ 2 \sqrt{2} x + 1 = 0 \\ x = \frac{ - 1}{2 \sqrt{2} }$

$x = \frac{3}{ \sqrt{2} } and \frac{ - 1}{2 \sqrt{2} }$

Answer 2

$\large\underline\purple{\bold{Solution :- }}$

$\tt \:  \longrightarrow \: {4x}^{2} + 5 \sqrt{2} x - 3$

$\tt \:  \longrightarrow \: {4x}^{2} + 6 \sqrt{2} x - \sqrt{2} x - 3$

$\tt \:  \longrightarrow \: 2 \times \sqrt{2} \times \sqrt{2} {x}^{2} - 3 \times 2 \times \sqrt{2} x - \sqrt{2} x - 3$

$\tt \:  \longrightarrow \: 2 \sqrt{2} x( \sqrt{2} x + 3) - 1( \sqrt{2} x + 3)$

$\tt \:  \longrightarrow \: ( \sqrt{2} x + 3)(2 \sqrt{2} x - 1)$

$\tt \:  \longrightarrow \: Hence \: zeroes \: are$

$\tt \:  \longrightarrow \: \sqrt{2} x + 3 = 0 \: or \: 2 \sqrt{2} x - 1 = 0$

$\tt \:  \longrightarrow \: x = - \dfrac{3}{ \sqrt{2} } \: or \: x \: = \dfrac{1}{2 \sqrt{2} }$