(B) To verify if $x = \frac{1}{2}$ is a solution,substitute the value into the quadratic equation $2x^{2} - 6x + 3 = 0$.Substitute $x = \frac{1}{2}$:$

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(B) To verify if $x = \frac{1}{2}$ is a solution,substitute the value into the quadratic equation $2x^{2} - 6x + 3 = 0$.Substitute $x = \frac{1}{2}$:$2(\frac{1}{2})^{2} - 6(\frac{1}{2}) + 3$$= 2(\frac{1}{4}) - 3 + 3$$= \frac{1}{2} - 3 + 3$$= \frac{1}{2}$Since the result is $\frac{1}{2}$ and not $0$,$x = \frac{1}{2}$ is not a solution to the given quadratic equation.

Class 10 Mathematics Quadratic Equations Mix Examples - Quadratic Equations

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Verify whether the given value of $x$ is a solution of the quadratic equation or not: $2x^{2} - 6x + 3 = 0$; $x = \frac{1}{2}$.

A
Yes
B
No
C
Cannot be determined
D
None of these

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Mix Examples - Quadratic Equations

State whether the quadratic equation $(x-1)(x+2)+2=0$ has two distinct real roots. Justify your answer.

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Every quadratic polynomial has at the most ......... zero/s.

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Find the discriminant of the following quadratic equation: $5x^{2} - 4\sqrt{2}x - 1 = 0$.

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If the following quadratic equation has two equal and real roots,then find the value of $k$: $x^{2} + 2kx + 4 = 0$.

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The roots of $4x^{2} + 2x + \frac{1}{4} = 0$ are $\ldots \ldots \ldots \ldots.$

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Verify whether the given value of $x$ is a solution of the quadratic equation or not: $2x^{2} - 6x + 3 = 0$; $x = \frac{1}{2}$.

Similar Questions

State whether the quadratic equation $(x-1)(x+2)+2=0$ has two distinct real roots. Justify your answer.

Every quadratic polynomial has at the most ......... zero/s.

Find the discriminant of the following quadratic equation: $5x^{2} - 4\sqrt{2}x - 1 = 0$.

If the following quadratic equation has two equal and real roots,then find the value of $k$: $x^{2} + 2kx + 4 = 0$.

The roots of $4x^{2} + 2x + \frac{1}{4} = 0$ are $\ldots \ldots \ldots \ldots.$

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