Joana solve the equation 4x^2 - 4x - 3 = 0. she followed the steps to solve a quadratic equation by grouping. However, institution is that the two groupings are not share common binomial, as shown in her work below. she concluded that the quadratic equation has no solution solutions. is Joana’s conclusion correct? Review Joana’s work to answer the question m.

Question

Step 1: 4x^2 + 2x + (-6x) - 3 = 0
Step 2: (4x^2 + 2x) + (-6x - 3) = 0
Step 3: x(4x + 2x) + (-3(2x + 1))= 0

Answer 1

Let's review Joana's steps to determine if her conclusion about the quadratic equation \( 4x^2 - 4x - 3 = 0 \) having no solutions is correct.

1. Starting equation: \[ 4x^2 - 4x - 3 = 0 \]

2. Step 1: Joana rewrote the equation as: \[ 4x^2 + 2x - 6x - 3 = 0 \] This step is correct as it still represents the same equation (the middle term \(-4x\) can be split into \(2x - 6x\)).

3. Step 2: Joana grouped the terms: \[ (4x^2 + 2x) + (-6x - 3) = 0 \] This is also a correct grouping.

4. Step 3: Joana factored each group: \[ x(4x + 2) - 3(2x + 1) = 0 \] This step contains an error. The factored forms are not presenting a common binomial correctly because \(4x + 2\) does not match with \(-3(2x + 1)\).

Resolving the equation using the quadratic formula:

The correct way to solve the quadratic equation is to apply the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] For our equation, where \(a = 4\), \(b = -4\), and \(c = -3\):

• Calculate the discriminant: \[ b^2 - 4ac = (-4)^2 - 4(4)(-3) = 16 + 48 = 64 \] Since the discriminant is positive (\(64\)), this means that there are two real solutions.

• Finally, applying the quadratic formula: \[ x = \frac{-(-4) \pm \sqrt{64}}{2(4)} = \frac{4 \pm 8}{8} \] This gives us two solutions:

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

Conclusion:

Joana's conclusion that the quadratic equation has no solutions is not correct. The equation \(4x^2 - 4x - 3 = 0\) does have two real solutions.