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(A) To verify if $x = -\sqrt{2}$ is a solution,substitute $x = -\sqrt{2}$ into the quadratic equation $\sqrt{2} x^{2} + 7 x + 5 \sqrt{2} = 0$.$LHS$ $=

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Yes,it is a solution.

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(A) To verify if $x = -\sqrt{2}$ is a solution,substitute $x = -\sqrt{2}$ into the quadratic equation $\sqrt{2} x^{2} + 7 x + 5 \sqrt{2} = 0$.$LHS$ $= \sqrt{2}(-\sqrt{2})^{2} + 7(-\sqrt{2}) + 5 \sqrt{2}$$= \sqrt{2}(2) - 7 \sqrt{2} + 5 \sqrt{2}$$= 2 \sqrt{2} - 7 \sqrt{2} + 5 \sqrt{2}$$= (2 - 7 + 5) \sqrt{2}$$= 0 	imes \sqrt{2} = 0$Since $LHS$ = $RHS$,$x = -\sqrt{2}$ is a solution of the given quadratic equation.

Verify whether the given value of $x$ is a solution of the quadratic equation or not: $\sqrt{2} x^{2}+7 x+5 \sqrt{2}=0, x=-\sqrt{2}$

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