We are given the quadratic equation:
$$
3x^2 + 4x - 15 = 0
$$
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 3, we divide every term by 3:
$$
x^2 + \frac{4}{3}x - 5 = 0
$$
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Step 2: Move the constant term to the other side
$$
x^2 + \frac{4}{3}x = 5
$$
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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 $x$: $\frac{4}{3}$
- Half of that: $\frac{2}{3}$
- Square of that: $\left(\frac{2}{3}\right)^2 = \frac{4}{9}$
Add $\frac{4}{9}$ to both sides:
$$
x^2 + \frac{4}{3}x + \frac{4}{9} = 5 + \frac{4}{9}
$$
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Step 4: Write the left side as a perfect square
$$
\left(x + \frac{2}{3}\right)^2 = \frac{45}{9} + \frac{4}{9} = \frac{49}{9}
$$
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Step 5: Take the square root of both sides
$$
x + \frac{2}{3} = \pm \sqrt{\frac{49}{9}} = \pm \frac{7}{3}
$$
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Step 6: Solve for $x$
$$
x = -\frac{2}{3} \pm \frac{7}{3}
$$
Now compute both solutions:
- $x = -\frac{2}{3} + \frac{7}{3} = \frac{5}{3}$
- $x = -\frac{2}{3} - \frac{7}{3} = -\frac{9}{3} = -3$
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✔ Final Answer:
$$
\boxed{x = \frac{5}{3} \quad \text{or} \quad x = -3}
$$
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Verification (Optional):
Plug back into original equation:
1. $x = \frac{5}{3}$:
$$
3\left(\frac{5}{3}\right)^2 + 4\left(\frac{5}{3}\right) - 15 = 3\cdot\frac{25}{9} + \frac{20}{3} - 15 = \frac{75}{9} + \frac{60}{9} - \frac{135}{9} = \frac{0}{9} = 0
$$
2. $x = -3$:
$$
3(-3)^2 + 4(-3) - 15 = 3(9) - 12 - 15 = 27 - 12 - 15 = 0
$$
✔ Both solutions check out.
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Let me know if you'd like a visual explanation or steps in a different format!
Parent Tip: Review the logic above to help your child master the concept of completing the square practice worksheet.
