Discover the best answers at Westonci.ca, where experts share their insights and knowledge with you. Explore thousands of questions and answers from a knowledgeable community of experts ready to help you find solutions. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.
Sagot :
To solve the equation [tex]\(4^P + \frac{1}{4^P} = \frac{257}{16}\)[/tex], we can follow these steps:
1. Substitute: Let [tex]\(4^P = x\)[/tex]. Then the equation becomes:
[tex]\[ x + \frac{1}{x} = \frac{257}{16} \][/tex]
2. Clear the fraction: Multiply both sides of the equation by [tex]\(x\)[/tex] to remove the fraction:
[tex]\[ x^2 + 1 = x \cdot \frac{257}{16} \][/tex]
3. Form a quadratic equation: Rearrange the terms to form a standard quadratic equation:
[tex]\[ x^2 + 1 = \frac{257}{16} x \][/tex]
Multiplying through by 16 to clear the fraction:
[tex]\[ 16x^2 + 16 = 257x \][/tex]
Rearrange this to standard quadratic form:
[tex]\[ 16x^2 - 257x + 16 = 0 \][/tex]
4. Solve the quadratic equation: Use the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 16\)[/tex], [tex]\(b = -257\)[/tex], and [tex]\(c = 16\)[/tex]:
Calculate the discriminant:
[tex]\[ \Delta = b^2 - 4ac = (-257)^2 - 4 \cdot 16 \cdot 16 = 66049 - 1024 = 65025 \][/tex]
Since the discriminant is positive, we have two real roots:
[tex]\[ x_{1,2} = \frac{257 \pm \sqrt{65025}}{32} \][/tex]
The square root of 65025 is 255. So we get:
[tex]\[ x_1 = \frac{257 + 255}{32} = \frac{512}{32} = 16 \][/tex]
[tex]\[ x_2 = \frac{257 - 255}{32} = \frac{2}{32} = \frac{1}{16} \][/tex]
5. Back-substitute: Recall that [tex]\(x = 4^P\)[/tex]:
For [tex]\(x_1 = 16\)[/tex]:
[tex]\[ 4^P = 16 \implies 4^P = 4^2 \implies P = 2 \][/tex]
For [tex]\(x_2 = \frac{1}{16}\)[/tex]:
[tex]\[ 4^P = \frac{1}{16} \implies 4^P = 4^{-2} \implies P = -2 \][/tex]
6. Conclusion: The values of [tex]\(P\)[/tex] that satisfy the equation [tex]\(4^P + \frac{1}{4^P} = \frac{257}{16}\)[/tex] are:
[tex]\[ P = 2 \quad \text{and} \quad P = -2 \][/tex]
1. Substitute: Let [tex]\(4^P = x\)[/tex]. Then the equation becomes:
[tex]\[ x + \frac{1}{x} = \frac{257}{16} \][/tex]
2. Clear the fraction: Multiply both sides of the equation by [tex]\(x\)[/tex] to remove the fraction:
[tex]\[ x^2 + 1 = x \cdot \frac{257}{16} \][/tex]
3. Form a quadratic equation: Rearrange the terms to form a standard quadratic equation:
[tex]\[ x^2 + 1 = \frac{257}{16} x \][/tex]
Multiplying through by 16 to clear the fraction:
[tex]\[ 16x^2 + 16 = 257x \][/tex]
Rearrange this to standard quadratic form:
[tex]\[ 16x^2 - 257x + 16 = 0 \][/tex]
4. Solve the quadratic equation: Use the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 16\)[/tex], [tex]\(b = -257\)[/tex], and [tex]\(c = 16\)[/tex]:
Calculate the discriminant:
[tex]\[ \Delta = b^2 - 4ac = (-257)^2 - 4 \cdot 16 \cdot 16 = 66049 - 1024 = 65025 \][/tex]
Since the discriminant is positive, we have two real roots:
[tex]\[ x_{1,2} = \frac{257 \pm \sqrt{65025}}{32} \][/tex]
The square root of 65025 is 255. So we get:
[tex]\[ x_1 = \frac{257 + 255}{32} = \frac{512}{32} = 16 \][/tex]
[tex]\[ x_2 = \frac{257 - 255}{32} = \frac{2}{32} = \frac{1}{16} \][/tex]
5. Back-substitute: Recall that [tex]\(x = 4^P\)[/tex]:
For [tex]\(x_1 = 16\)[/tex]:
[tex]\[ 4^P = 16 \implies 4^P = 4^2 \implies P = 2 \][/tex]
For [tex]\(x_2 = \frac{1}{16}\)[/tex]:
[tex]\[ 4^P = \frac{1}{16} \implies 4^P = 4^{-2} \implies P = -2 \][/tex]
6. Conclusion: The values of [tex]\(P\)[/tex] that satisfy the equation [tex]\(4^P + \frac{1}{4^P} = \frac{257}{16}\)[/tex] are:
[tex]\[ P = 2 \quad \text{and} \quad P = -2 \][/tex]
Answer:
P = 2 and P = -2 .
Step-by-step explanation:
Let's solve the equation:
[tex]4^P + \frac{1}{4^P} = 16 + \frac{1}{16}[/tex]
First, simplify the right-hand side:
[tex]16 + \frac{1}{16} = 16 + \frac{1}{16} = \frac{256}{16} + \frac{1}{16} = \frac{256 + 1}{16} = \frac{257}{16}[/tex]
So the equation becomes:
[tex]4^P + \frac{1}{4^P} = \frac{257}{16}[/tex]
Let x = [tex]4^P[/tex] . Then the equation is:
[tex]x + \frac{1}{x} = \frac{257}{16}[/tex]
Multiply both sides by 16x to clear the fraction:
[tex]16x^2 + 16 = 257x[/tex]
Rearrange to form a quadratic equation:
[tex]16x^2 - 257x + 16 = 0[/tex]
Solve this quadratic equation using the quadratic formula [tex]x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} :[/tex]
Here, a = 16 , b = -257 , and c = 16 .
[tex]x = \frac{-(-257) \pm \sqrt{(-257)^2 - 4 \cdot 16 \cdot 16}}{2 \cdot 16} \\\\ x = \frac{257 \pm \sqrt{66049 - 1024}}{32} \\\\ x = \frac{257 \pm \sqrt{65025}}{32} \\\\ x = \frac{257 \pm 255}{32} \\\\[/tex]
So, we have two solutions:
[tex]x = \frac{257 + 255}{32} = \frac{512}{32} = 16 \\\\ x = \frac{257 - 255}{32} = \frac{2}{32} = \frac{1}{16}[/tex]
Thus, x = 16 or x = [tex]\frac{1}{16} .[/tex]
Since x = [tex]4^P :[/tex]
For x = 16 :
[tex]4^P = 16 \\\\ (2^2)^P = 2^4 \\\\ 2^{2P} = 2^4 \\\\ 2P = 4 \\\\ P = 2[/tex]
[tex]For x = \frac{1}{16} :\\\\ 4^P = \frac{1}{16} \\\\ (2^2)^P = 2^{-4} \\\\ 2^{2P} = 2^{-4} \\\\ 2P = -4 \\\\ P = -2 \\\\[/tex]
So, the solutions are P = 2 and P = -2 .
We hope our answers were helpful. Return anytime for more information and answers to any other questions you may have. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Thank you for visiting Westonci.ca. Stay informed by coming back for more detailed answers.
