Question

$\sqrt{2} x ^{2} - + 7x + 5 \sqrt{2} = 0$
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Answer 1

$\huge{\boxed{\red{\star\;Answer}}}$

$\large{\underline{\blue{\star\;Note}}}$

• If $ax^{2}+bx+c$ is a quadratic equation then the roots of quadratic equation are given by ,

• $x\;=\;-b+\;or-\;\dfrac{\sqrt{b^{2}-4ac}}{2a}$

$\underline{\pink{Given\;equation\;\sqrt{2}x^{2}+7x+5\sqrt{2}}}$

• Using the above formula, by Comparing the terms,
• a = $\sqrt{2}$
• b = 7
• c = $5(\sqrt{2})$

$\underline{\pink{Substituting\;in\;formula}}$

• $x\;=\;-b+\;or-\;\dfrac{\sqrt{b^{2}-4ac}}{2a}$

• $x\;=\;-7+\;or-\;\dfrac{\sqrt{7^{2}-4(\sqrt{2})(5\sqrt{2})}}{2(\sqrt{2})}$

• $x\;=\;-7+\;or-\;\dfrac{\sqrt{49-40}}{2(\sqrt{2})}$

• $x\;=\;-7+\;or-\;\dfrac{\sqrt{9}}{2(\sqrt{2})}$

• $x\;=\;-7+\;or-\;\dfrac{3}{2\sqrt{2}}$

$\boxed{\red{The\; required\;values\;of\;x\;are\;x=-7+\dfrac{3}{2\sqrt{2}}\;and\;x=-7-\dfrac{3}{2\sqrt{2}}}}$