Question

joana solved the equation 4x^2-4x-3=0. she followed the steps to solve a quadratic equation by grouping. however, in step 3 she noticed that the two groupings did not share a common binomial, as shown is her work below. she concluded that the quadratic equation 4x^2-4x-3=0 has no solutions. is joanna's conclusion correct review joanna's work to answer the question

Answer 1

To determine if Joana's conclusion is correct, we need to solve the quadratic equation \( 4x^2 - 4x - 3 = 0 \) and examine her reasoning.

To solve the quadratic equation, we can either factor it or use the quadratic formula. Here, we will use the quadratic formula, which is given by:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

In our equation \( 4x^2 - 4x - 3 = 0 \), the coefficients are:
- \( a = 4 \)
- \( b = -4 \)
- \( c = -3 \)

Let's calculate the discriminant \( b^2 - 4ac \):

\[
b^2 - 4ac = (-4)^2 - 4 \cdot 4 \cdot (-3) = 16 + 48 = 64
\]

Since the discriminant is positive (\( 64 > 0 \)), there are two distinct real solutions. Now, let's substitute into the quadratic formula:

\[
x = \frac{-(-4) \pm \sqrt{64}}{2 \cdot 4} = \frac{4 \pm 8}{8}
\]

This gives us two solutions:

1. \( x = \frac{12}{8} = \frac{3}{2} \)
2. \( x = \frac{-4}{8} = -\frac{1}{2} \)

Thus, the solutions are \( x = \frac{3}{2} \) and \( x = -\frac{1}{2} \).

### Conclusion

Since there are indeed solutions to the quadratic equation \( 4x^2 - 4x - 3 = 0 \), Joana's conclusion that the equation has no solutions is **incorrect**. It's important to note that the absence of a common binomial factor does not imply that the equation has no solutions; rather, it indicates that the quadratic might not be factorable in nice integers or rational numbers, and that's why using the quadratic formula or completing the square can be effective alternatives.