Answer :
Let's solve the quadratic equation step-by-step, using the quadratic equation provided:
[tex]\[ x^2 - x - \frac{3}{4} = 0 \][/tex]
First, we'll rewrite the equation by moving \(-\frac{3}{4}\) to the other side:
[tex]\[ x^2 - x = \frac{3}{4} \][/tex]
Next, to make this suitable for completing the square, we add \(\left(\frac{1}{2}\right)^2\) to both sides. This is because the coefficient of \(x\) is \(-1\), and half of \(-1\) is \(-\frac{1}{2}\), so when squared, it is \(\left(\frac{1}{2}\right)^2\):
[tex]\[ x^2 - x + \left(\frac{1}{2}\right)^2 = \frac{3}{4} + \left(\frac{1}{2}\right)^2 \][/tex]
Calculating the right-hand side:
[tex]\[ x^2 - x + \frac{1}{4} = \frac{3}{4} + \frac{1}{4} \][/tex]
[tex]\[ x^2 - x + \frac{1}{4} = 1 \][/tex]
The left-hand side is now a perfect square trinomial:
[tex]\[ (x - \frac{1}{2})^2 = 1 \][/tex]
To find \(x\), we take the square root of both sides:
[tex]\[ x - \frac{1}{2} = \pm 1 \][/tex]
This gives us two equations to solve:
[tex]\[ x - \frac{1}{2} = 1 \quad \text{or} \quad x - \frac{1}{2} = -1 \][/tex]
Solving these:
[tex]\[ x = 1 + \frac{1}{2} = \frac{3}{2} \][/tex]
[tex]\[ x = -1 + \frac{1}{2} = -\frac{1}{2} \][/tex]
Therefore, the solutions to the quadratic equation \(x^2 - x - \frac{3}{4} = 0\) are:
[tex]\[ x = \frac{3}{2} \quad \text{and} \quad x = -\frac{1}{2} \][/tex]
From the given possible solutions:
- \(-\frac{1}{4}\)
- \(\frac{1}{2}\)
- \(\frac{3}{2}\)
- \(\frac{3}{4}\)
The correct solution provided by the options is \(\frac{3}{2}\).
Thus, the solution of \(x^2 - x - \frac{3}{4} = 0\) from the options given is:
[tex]\[ \boxed{\frac{3}{2}} \][/tex]
[tex]\[ x^2 - x - \frac{3}{4} = 0 \][/tex]
First, we'll rewrite the equation by moving \(-\frac{3}{4}\) to the other side:
[tex]\[ x^2 - x = \frac{3}{4} \][/tex]
Next, to make this suitable for completing the square, we add \(\left(\frac{1}{2}\right)^2\) to both sides. This is because the coefficient of \(x\) is \(-1\), and half of \(-1\) is \(-\frac{1}{2}\), so when squared, it is \(\left(\frac{1}{2}\right)^2\):
[tex]\[ x^2 - x + \left(\frac{1}{2}\right)^2 = \frac{3}{4} + \left(\frac{1}{2}\right)^2 \][/tex]
Calculating the right-hand side:
[tex]\[ x^2 - x + \frac{1}{4} = \frac{3}{4} + \frac{1}{4} \][/tex]
[tex]\[ x^2 - x + \frac{1}{4} = 1 \][/tex]
The left-hand side is now a perfect square trinomial:
[tex]\[ (x - \frac{1}{2})^2 = 1 \][/tex]
To find \(x\), we take the square root of both sides:
[tex]\[ x - \frac{1}{2} = \pm 1 \][/tex]
This gives us two equations to solve:
[tex]\[ x - \frac{1}{2} = 1 \quad \text{or} \quad x - \frac{1}{2} = -1 \][/tex]
Solving these:
[tex]\[ x = 1 + \frac{1}{2} = \frac{3}{2} \][/tex]
[tex]\[ x = -1 + \frac{1}{2} = -\frac{1}{2} \][/tex]
Therefore, the solutions to the quadratic equation \(x^2 - x - \frac{3}{4} = 0\) are:
[tex]\[ x = \frac{3}{2} \quad \text{and} \quad x = -\frac{1}{2} \][/tex]
From the given possible solutions:
- \(-\frac{1}{4}\)
- \(\frac{1}{2}\)
- \(\frac{3}{2}\)
- \(\frac{3}{4}\)
The correct solution provided by the options is \(\frac{3}{2}\).
Thus, the solution of \(x^2 - x - \frac{3}{4} = 0\) from the options given is:
[tex]\[ \boxed{\frac{3}{2}} \][/tex]