Verify whether the given value of $x$ is a solution of the quadratic equation or not: $2x^2 - 5x + 3 = 0$; $x = \frac{1}{2}$.
(B) To verify if $x = \frac{1}{2}$ is a solution,substitute $x = \frac{1}{2}$ into the quadratic equation $2x^2 - 5x + 3 = 0$.
$LHS$ $= 2(\frac{1}{2})^2 - 5(\frac{1}{2}) + 3$
$= 2(\frac{1}{4}) - \frac{5}{2} + 3$
$= \frac{1}{2} - \frac{5}{2} + 3$
$= -\frac{4}{2} + 3$
$= -2 + 3 = 1$
Since $LHS$ $\neq$ $RHS$ (where $RHS$ $= 0$),$x = \frac{1}{2}$ is not a solution of the given quadratic equation.
$LHS$ $= 2(\frac{1}{2})^2 - 5(\frac{1}{2}) + 3$
$= 2(\frac{1}{4}) - \frac{5}{2} + 3$
$= \frac{1}{2} - \frac{5}{2} + 3$
$= -\frac{4}{2} + 3$
$= -2 + 3 = 1$
Since $LHS$ $\neq$ $RHS$ (where $RHS$ $= 0$),$x = \frac{1}{2}$ is not a solution of the given quadratic equation.
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