Looking for trustworthy answers? Westonci.ca is the ultimate Q&A platform where experts share their knowledge on various topics. Our platform offers a seamless experience for finding reliable answers from a network of experienced professionals. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly 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.
Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Westonci.ca is here to provide the answers you seek. Return often for more expert solutions.