Answer :
To solve the quadratic equation [tex]\(4x^2 + 6x = 4\)[/tex], let's proceed step by step.
1. Rewrite the equation in standard form:
The equation is [tex]\(4x^2 + 6x = 4\)[/tex]. We need to move all terms to one side to set the equation equal to zero.
[tex]\[ 4x^2 + 6x - 4 = 0 \][/tex]
2. Identify the coefficients:
For a quadratic equation in the form [tex]\(ax^2 + bx + c = 0\)[/tex]:
- [tex]\(a = 4\)[/tex]
- [tex]\(b = 6\)[/tex]
- [tex]\(c = -4\)[/tex]
3. Use the quadratic formula to find the roots:
The quadratic formula is given by:
[tex]\[ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \][/tex]
Plugging in the coefficients [tex]\(a = 4\)[/tex], [tex]\(b = 6\)[/tex], and [tex]\(c = -4\)[/tex]:
[tex]\[ x = \frac{{-6 \pm \sqrt{{6^2 - 4 \cdot 4 \cdot (-4)}}}}{2 \cdot 4} \][/tex]
Simplify inside the square root:
[tex]\[ x = \frac{{-6 \pm \sqrt{{36 + 64}}}}{8} \][/tex]
[tex]\[ x = \frac{{-6 \pm \sqrt{{100}}}}{8} \][/tex]
Since [tex]\(\sqrt{100} = 10\)[/tex], we can continue:
[tex]\[ x = \frac{{-6 \pm 10}}{8} \][/tex]
4. Calculate the two possible solutions:
- For the positive root:
[tex]\[ x = \frac{{-6 + 10}}{8} = \frac{4}{8} = 0.5 \][/tex]
- For the negative root:
[tex]\[ x = \frac{{-6 - 10}}{8} = \frac{-16}{8} = -2 \][/tex]
5. Conclusion:
The solutions to the quadratic equation [tex]\(4x^2 + 6x - 4 = 0\)[/tex] are:
[tex]\[ x = 0.5 \quad \text{and} \quad x = -2 \][/tex]
Therefore, the correct answers to the question are [tex]\(0.5\)[/tex] and [tex]\(-2\)[/tex]. The other given options [tex]\(0.75\)[/tex], [tex]\(-1.5\)[/tex], [tex]\(-2.25\)[/tex], [tex]\(0\)[/tex], and "No real solutions" are incorrect.
1. Rewrite the equation in standard form:
The equation is [tex]\(4x^2 + 6x = 4\)[/tex]. We need to move all terms to one side to set the equation equal to zero.
[tex]\[ 4x^2 + 6x - 4 = 0 \][/tex]
2. Identify the coefficients:
For a quadratic equation in the form [tex]\(ax^2 + bx + c = 0\)[/tex]:
- [tex]\(a = 4\)[/tex]
- [tex]\(b = 6\)[/tex]
- [tex]\(c = -4\)[/tex]
3. Use the quadratic formula to find the roots:
The quadratic formula is given by:
[tex]\[ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \][/tex]
Plugging in the coefficients [tex]\(a = 4\)[/tex], [tex]\(b = 6\)[/tex], and [tex]\(c = -4\)[/tex]:
[tex]\[ x = \frac{{-6 \pm \sqrt{{6^2 - 4 \cdot 4 \cdot (-4)}}}}{2 \cdot 4} \][/tex]
Simplify inside the square root:
[tex]\[ x = \frac{{-6 \pm \sqrt{{36 + 64}}}}{8} \][/tex]
[tex]\[ x = \frac{{-6 \pm \sqrt{{100}}}}{8} \][/tex]
Since [tex]\(\sqrt{100} = 10\)[/tex], we can continue:
[tex]\[ x = \frac{{-6 \pm 10}}{8} \][/tex]
4. Calculate the two possible solutions:
- For the positive root:
[tex]\[ x = \frac{{-6 + 10}}{8} = \frac{4}{8} = 0.5 \][/tex]
- For the negative root:
[tex]\[ x = \frac{{-6 - 10}}{8} = \frac{-16}{8} = -2 \][/tex]
5. Conclusion:
The solutions to the quadratic equation [tex]\(4x^2 + 6x - 4 = 0\)[/tex] are:
[tex]\[ x = 0.5 \quad \text{and} \quad x = -2 \][/tex]
Therefore, the correct answers to the question are [tex]\(0.5\)[/tex] and [tex]\(-2\)[/tex]. The other given options [tex]\(0.75\)[/tex], [tex]\(-1.5\)[/tex], [tex]\(-2.25\)[/tex], [tex]\(0\)[/tex], and "No real solutions" are incorrect.
