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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 .
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