Answer :

GinnyAnswer
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]
Hi1315

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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