To solve the quadratic equation [tex]\(4x^2 - 7x = -3\)[/tex], let's first rearrange it into standard form:
[tex]\[
4x^2 - 7x + 3 = 0
\][/tex]
Next, we’ll use the quadratic formula to find the roots of the equation [tex]\(ax^2 + bx + c = 0\)[/tex]. The quadratic formula is given by:
[tex]\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
For the given equation [tex]\(4x^2 - 7x + 3 = 0\)[/tex], the coefficients are:
- [tex]\(a = 4\)[/tex]
- [tex]\(b = -7\)[/tex]
- [tex]\(c = 3\)[/tex]
Now, let's substitute these values into the quadratic formula:
[tex]\[
x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4 \cdot 4 \cdot 3}}{2 \cdot 4}
\][/tex]
First, calculate the discriminant ([tex]\(b^2 - 4ac\)[/tex]):
[tex]\[
b^2 - 4ac = (-7)^2 - 4 \cdot 4 \cdot 3 = 49 - 48 = 1
\][/tex]
Now, substitute the discriminant back into the quadratic formula:
[tex]\[
x = \frac{7 \pm \sqrt{1}}{8}
\][/tex]
Since [tex]\(\sqrt{1} = 1\)[/tex], we can simplify this further:
[tex]\[
x = \frac{7 \pm 1}{8}
\][/tex]
We now have two potential solutions:
[tex]\[
x_1 = \frac{7 + 1}{8} = \frac{8}{8} = 1
\][/tex]
[tex]\[
x_2 = \frac{7 - 1}{8} = \frac{6}{8} = \frac{3}{4}
\][/tex]
Thus, the solutions to the equation [tex]\(4x^2 - 7x + 3 = 0\)[/tex] are:
[tex]\[
x = \left\{ \frac{3}{4}, 1 \right\}
\][/tex]
So, the correct answer is:
A. [tex]\(x = \left\{ \frac{3}{4}, 1 \right\} \)[/tex]
