How To Solve Quadratic Equations By Completing The Square - Free Printable

We are given the quadratic equation:

$$
2x^2 - 7x + 3 = 0
$$

We are to solve this by completing the square.

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Step 1: Divide by the leading coefficient (to make the coefficient of $x^2$ equal to 1)

Since the coefficient of $x^2$ is 2, we divide every term by 2:

$$
x^2 - \frac{7}{2}x + \frac{3}{2} = 0
$$

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Step 2: Move the constant term to the other side

$$
x^2 - \frac{7}{2}x = -\frac{3}{2}
$$

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Step 3: Complete the square

To complete the square, take half of the coefficient of $x$, square it, and add it to both sides.

- Coefficient of $x$: $-\frac{7}{2}$
- Half of that: $-\frac{7}{4}$
- Square of that: $\left(-\frac{7}{4}\right)^2 = \frac{49}{16}$

Add $\frac{49}{16}$ to both sides:

$$
x^2 - \frac{7}{2}x + \frac{49}{16} = -\frac{3}{2} + \frac{49}{16}
$$

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Step 4: Simplify both sides

Left side becomes a perfect square:

$$
\left(x - \frac{7}{4}\right)^2 = -\frac{3}{2} + \frac{49}{16}
$$

Convert $-\frac{3}{2}$ to sixteenths:

$$
-\frac{3}{2} = -\frac{24}{16}
$$

So:

$$
\left(x - \frac{7}{4}\right)^2 = -\frac{24}{16} + \frac{49}{16} = \frac{25}{16}
$$

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Step 5: Take the square root of both sides

$$
x - \frac{7}{4} = \pm \sqrt{\frac{25}{16}} = \pm \frac{5}{4}
$$

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Step 6: Solve for $x$

$$
x = \frac{7}{4} \pm \frac{5}{4}
$$

So:

- $x = \frac{7}{4} + \frac{5}{4} = \frac{12}{4} = 3$
- $x = \frac{7}{4} - \frac{5}{4} = \frac{2}{4} = \frac{1}{2}$

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✔ Final Answer:

$$
\boxed{x = 3 \quad \text{or} \quad x = \frac{1}{2}}
$$

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Verification (Optional):

Plug back into original equation:

For $x = 3$:
$2(3)^2 - 7(3) + 3 = 2(9) - 21 + 3 = 18 - 21 + 3 = 0$ ✔

For $x = \frac{1}{2}$:
$2\left(\frac{1}{2}\right)^2 - 7\left(\frac{1}{2}\right) + 3 = 2\left(\frac{1}{4}\right) - \frac{7}{2} + 3 = \frac{1}{2} - \frac{7}{2} + 3 = -3 + 3 = 0$ ✔

Both solutions work!

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Summary:

We solved $2x^2 - 7x + 3 = 0$ by completing the square, resulting in:

$$
\boxed{x = 3 \quad \text{or} \quad x = \frac{1}{2}}
$$