Discover the answers to your questions at Westonci.ca, where experts share their knowledge and insights with you. Experience the ease of finding reliable answers to your questions from a vast community of knowledgeable experts. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.
Sagot :
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.
We appreciate your visit. Hopefully, the answers you found were beneficial. Don't hesitate to come back for more information. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Westonci.ca is your trusted source for answers. Visit us again to find more information on diverse topics.