Question

find the roots of √2x^2-x-√2=0 are​

Answer 1

Answer:

x= root 2

Step-by-step explanation:

after solving square

2x-x-root2=0

x=root2

Answer 2

⠀⠀⠀⠀⠀${ \huge \bf{ \mid{ \overline{ \underline{ \pink{QUESTION}}}} \mid}} \longrightarrow</p><p>$

⠀⠀find the roots of √2x^2-x-√2=0 .

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⠀⠀⠀⠀⠀⠀$\huge{ \underline{ \red{ \bold{ \underline{ \bf{AnSweR }}}}}}$

⠀⠀$\large\underline{ \underline{ \blue{ \bold {Given}}}} = >$

quadratic equation =$\bf \sqrt{2} {x}^{2} - x - \sqrt{2} = 0$

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⠀⠀$\large\underline{ \underline{ \blue{ \bold {To \:Find}}}} = >$

we have to find the roots .

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⠀⠀⠀⠀⠀✾$\huge\underline{ \underline{ \green{ \bold{solution}}}}$

⠀ By using quadratic formula

⠀⠀⠀$\boxed{ \fbox{ \red{ \frac{ - b + - \sqrt{b {}^{2} - 4ac } }{2a} }}}$

$= \frac{ - ( - 1) + - \sqrt{ - 1 {}^{2} - 4 \sqrt{2} \times ( - \sqrt{2)} } }{2 \sqrt{2} } \\ \\ = > \frac{1 + - \sqrt{1 + 4 \sqrt{2 \times \sqrt{2} } } }{2 \sqrt{2} } \\ \\ = > \frac{1 + - \sqrt{1 + 4 \times 2} }{2 \sqrt{2} } \\ \\ = > \frac{1 + - \sqrt{1 + 8} }{2 \sqrt{2} }$

⠀⠀⠀$\boxed{ \blue{ \fbox{ \green{x = \sqrt{2} \ \: \: \ \: :{ \pink { or}} \: \: \: \: \: { \red{ \: x = \frac{ - \sqrt{2} }{2} }}}}}}$

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hops this may help you

⠀⠀⠀⠀⠀⠀⠀⠀$\huge{ \pink{ \ddot{ \smile}}}$

⠀⠀⠀⠀⠀⠀$\huge \mathfrak{ \blue { \bigstar{thanks}}}$