Question

Find the discriminant of the quadratic equation
$\sqrt{2} x {}^{2} + 7x + 5 \sqrt{2 } = 0$
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Answer 1

Answer:

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Answer 2

Answer:

$\huge{\boxed{\sf{D=9}}}$

$\huge{\boxed{\sf{ROOTS\:OF\:THIS\:EQN=-\sqrt{2}\:AND\frac{-5}{\sqrt{2}}}}}$

Step-by-step explanation:

$\huge{\boxed{\sf{SOLUTION-}}}$

$\huge{\boxed{\sf{METHOD-}}}$

$\huge{\boxed{\sf{QUADRATIC\:FORMULA}}}$

$\sqrt{2} {x}^{2} + 7x + 5 \sqrt{2} = 0 \\ d = {b}^{2} - 4ac \\ d = {7}^{2} - 4( \sqrt{2} \times 5 \sqrt{2} ) \\ d = 49 - 4 \times 10 \\ d = 49 - 40 \\ d = 9 \\ again \\ x = \frac{ - b + \sqrt{d} }{2a} \\x = \frac{ - 7 + \sqrt{9} }{2 \sqrt{2} } \\ x = \frac{ - 7 + 3}{2 \sqrt{2} } \\ x = \frac{ - 4}{2 \sqrt{2} } \\ x = - \sqrt{2} - - - - - 1st \: root \\ x = \frac{ - b - \sqrt{d} }{2a} \\ x = \frac{ - 7 - 3}{2 \sqrt{2} } \\ x = \frac{ - 10}{2 \sqrt{2} } \\ x = \frac{ - 5}{ \sqrt{2} } - - - - 2nd \: root$

$\huge{\boxed{\sf{D=9}}}$

$\huge{\boxed{\sf{ROOTS\:OF\:THIS\:EQN=-\sqrt{2}\:AND\frac{-5}{\sqrt{2}}}}}$