Answer :
To solve the quadratic equation [tex]\(3x^2 - 7x + 2 = 0\)[/tex] by completing the square, follow these steps:
1. Normalize the coefficient of [tex]\(x^2\)[/tex]:
The coefficient of [tex]\(x^2\)[/tex] is 3. To make it 1, divide every term in the equation by 3:
[tex]\[ \frac{3x^2}{3} - \frac{7x}{3} + \frac{2}{3} = 0 \implies x^2 - \frac{7}{3}x + \frac{2}{3} = 0 \][/tex]
2. Move the constant term to the right side of the equation:
[tex]\[ x^2 - \frac{7}{3}x = -\frac{2}{3} \][/tex]
3. Complete the square:
To complete the square, we need to add and subtract the square of half the coefficient of [tex]\(x\)[/tex]. The coefficient of [tex]\(x\)[/tex] is [tex]\(\frac{-7}{3}\)[/tex], so half of this coefficient is [tex]\(\frac{-7}{6}\)[/tex]. When squared, [tex]\(\left(\frac{-7}{6}\right)^2 = \frac{49}{36}\)[/tex].
Add and subtract [tex]\(\frac{49}{36}\)[/tex] to the left side of the equation:
[tex]\[ x^2 - \frac{7}{3}x + \frac{49}{36} = \frac{49}{36} - \frac{2}{3} \][/tex]
4. Simplify the right side:
Convert [tex]\(-\frac{2}{3}\)[/tex] to a fraction with a denominator of 36:
[tex]\[ \frac{2}{3} = \frac{24}{36} \][/tex]
So,
[tex]\[ \frac{49}{36} - \frac{24}{36} = \frac{25}{36} \][/tex]
Now the equation is:
[tex]\[ x^2 - \frac{7}{3}x + \frac{49}{36} = \frac{25}{36} \][/tex]
5. Rewrite the left side as a perfect square:
[tex]\[ \left(x - \frac{7}{6}\right)^2 = \frac{25}{36} \][/tex]
6. Solve the equation by taking the square root of both sides:
[tex]\[ x - \frac{7}{6} = \pm \frac{5}{6} \][/tex]
This gives us two equations to solve:
[tex]\[ x - \frac{7}{6} = \frac{5}{6} \quad \text{and} \quad x - \frac{7}{6} = -\frac{5}{6} \][/tex]
7. Solve for [tex]\(x\)[/tex]:
[tex]\[ x = \frac{7}{6} + \frac{5}{6} = \frac{12}{6} = 2 \][/tex]
[tex]\[ x = \frac{7}{6} - \frac{5}{6} = \frac{2}{6} = \frac{1}{3} \][/tex]
Thus, the solutions to the equation [tex]\(3x^2 - 7x + 2 = 0\)[/tex] are:
[tex]\[ x = 2 \quad \text{and} \quad x = \frac{1}{3} \][/tex]
In decimal form, the solutions are:
[tex]\[ x = 2.0 \quad \text{and} \quad x \approx 0.333 \][/tex]
1. Normalize the coefficient of [tex]\(x^2\)[/tex]:
The coefficient of [tex]\(x^2\)[/tex] is 3. To make it 1, divide every term in the equation by 3:
[tex]\[ \frac{3x^2}{3} - \frac{7x}{3} + \frac{2}{3} = 0 \implies x^2 - \frac{7}{3}x + \frac{2}{3} = 0 \][/tex]
2. Move the constant term to the right side of the equation:
[tex]\[ x^2 - \frac{7}{3}x = -\frac{2}{3} \][/tex]
3. Complete the square:
To complete the square, we need to add and subtract the square of half the coefficient of [tex]\(x\)[/tex]. The coefficient of [tex]\(x\)[/tex] is [tex]\(\frac{-7}{3}\)[/tex], so half of this coefficient is [tex]\(\frac{-7}{6}\)[/tex]. When squared, [tex]\(\left(\frac{-7}{6}\right)^2 = \frac{49}{36}\)[/tex].
Add and subtract [tex]\(\frac{49}{36}\)[/tex] to the left side of the equation:
[tex]\[ x^2 - \frac{7}{3}x + \frac{49}{36} = \frac{49}{36} - \frac{2}{3} \][/tex]
4. Simplify the right side:
Convert [tex]\(-\frac{2}{3}\)[/tex] to a fraction with a denominator of 36:
[tex]\[ \frac{2}{3} = \frac{24}{36} \][/tex]
So,
[tex]\[ \frac{49}{36} - \frac{24}{36} = \frac{25}{36} \][/tex]
Now the equation is:
[tex]\[ x^2 - \frac{7}{3}x + \frac{49}{36} = \frac{25}{36} \][/tex]
5. Rewrite the left side as a perfect square:
[tex]\[ \left(x - \frac{7}{6}\right)^2 = \frac{25}{36} \][/tex]
6. Solve the equation by taking the square root of both sides:
[tex]\[ x - \frac{7}{6} = \pm \frac{5}{6} \][/tex]
This gives us two equations to solve:
[tex]\[ x - \frac{7}{6} = \frac{5}{6} \quad \text{and} \quad x - \frac{7}{6} = -\frac{5}{6} \][/tex]
7. Solve for [tex]\(x\)[/tex]:
[tex]\[ x = \frac{7}{6} + \frac{5}{6} = \frac{12}{6} = 2 \][/tex]
[tex]\[ x = \frac{7}{6} - \frac{5}{6} = \frac{2}{6} = \frac{1}{3} \][/tex]
Thus, the solutions to the equation [tex]\(3x^2 - 7x + 2 = 0\)[/tex] are:
[tex]\[ x = 2 \quad \text{and} \quad x = \frac{1}{3} \][/tex]
In decimal form, the solutions are:
[tex]\[ x = 2.0 \quad \text{and} \quad x \approx 0.333 \][/tex]