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(A) To verify if $x = \frac{\sqrt{3}}{4}$ is a solution,substitute the value of $x$ into the quadratic equation $4 \sqrt{3} x^{2} + 5 x - 2 \sqrt{3} =

Vedclass · Accepted Answer

Yes,it is a solution.

Vedclass · Answer

(A) To verify if $x = \frac{\sqrt{3}}{4}$ is a solution,substitute the value of $x$ into the quadratic equation $4 \sqrt{3} x^{2} + 5 x - 2 \sqrt{3} = 0$.$LHS$ = $4 \sqrt{3} \left( \frac{\sqrt{3}}{4} \right)^{2} + 5 \left( \frac{\sqrt{3}}{4} \right) - 2 \sqrt{3}$$LHS$ = $4 \sqrt{3} \left( \frac{3}{16} \right) + \frac{5 \sqrt{3}}{4} - 2 \sqrt{3}$$LHS$ = $\frac{3 \sqrt{3}}{4} + \frac{5 \sqrt{3}}{4} - 2 \sqrt{3}$$LHS$ = $\frac{8 \sqrt{3}}{4} - 2 \sqrt{3}$$LHS$ = $2 \sqrt{3} - 2 \sqrt{3} = 0$Since $LHS$ = $RHS$,the given value $x = \frac{\sqrt{3}}{4}$ is a solution of the quadratic equation.

Verify whether the given value of $x$ is a solution of the quadratic equation or not: $4 \sqrt{3} x^{2} + 5 x - 2 \sqrt{3} = 0$; $x = \frac{\sqrt{3}}{4}$.

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