Question

Quadratic Equation class 10 ​

Answer 1

Step-by-step explanation:

by quadratic equation we get 2 answers 2root 2 and second solution is -root2 /2

Answer 2

$\huge\rm\underline\purple{SOLUTION:}$

$\sqrt{2} {x}^{2} - 3x - 2 \sqrt{2} = 0 \\ \\ the \: general \: form \: of \: a \: quadratic \: equation \: is \\ \\ ax {}^{2} + bx + c = 0 \\ \\ and \: x = \frac{ - b + \sqrt{ {b}^{2} - 4ac } }{2a} (or)x = \frac{ - b - \sqrt{ {b}^{2} - 4ac} }{2a} \\ \\ conparing \: the \: equation \: with \: general \: form \: of \: \\ equation \: we \: get \\ \\ a = \sqrt{2} \: and \: b = 3 \: and \: c = - 2 \sqrt{2} \\ \\ = > x = \frac{ - 3 + \sqrt{9 - 4( \sqrt{2})( - 2 \sqrt{2} ) } }{2( \sqrt{2)} } \: (or) \\ \\ x = \frac{ - 3 - \sqrt{9 - 4( \sqrt{2} )( - 2 \sqrt{2)} } }{2( \sqrt{2} )} \\ \\ = > x = \frac{ - 3 \sqrt{9 - 4( - 4)} }{2 \sqrt{2} } (or) \\ \\ x = \frac{ - 3 - \sqrt{9 - 4( - 4)} }{2 \sqrt{2} } \\ \\ = > x = \frac{ - 3 \sqrt{9 + 16} }{2 \sqrt{2} } \: (or) \: x = \frac{ - 3 - \sqrt{9 + 16} }{2 \sqrt{2} } \\ \\ = > x = \frac{ - 3 + \sqrt{25} }{2 \sqrt{2} } \: (or) \: x = \frac{ - 3 - \sqrt{25} }{2 \sqrt{2} } \\ \\ = > x = \frac{ - 3 + 5}{2 \sqrt{2} } \: (or) \: x = \frac{ - 3 - 5}{2 \sqrt{2} } \\ \\ = > x = \frac{2}{2 \sqrt{2} } \: (or) \: x = \frac{ - 8}{2 \sqrt{2} } \\ \\ = > x = \frac{1}{ \sqrt{2} } \: (or) \: x = \frac{ - 4}{ \sqrt{2} } \\ \\ rationalizing \: the \: denominator \\ \\ = > x = \frac{ \sqrt{2} }{2} \: (or) \: x = - 2 \sqrt{2}$