Question

11.1 2) x4 – 8x2 - 9 = 0; – 0 4) 16x4 - 409x2 + 225 = 0.​

Answer 1

Ответ:

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Answer 2

It seems like you have two quadratic equations that need to be solved. Let's solve each of them separately:

1. x^4 - 8x^2 - 9 = 0

To solve this equation, we can make a substitution to simplify it. Let u = x^2, then the equation becomes:

u^2 - 8u - 9 = 0

Now, we can solve this quadratic equation for u:

(u - 9)(u + 1) = 0

Setting each factor equal to zero:

u - 9 = 0 or u + 1 = 0

Solving for u:

u = 9 or u = -1

Now, we need to back-substitute to find the values of x:

For u = 9: x^2 = 9 x = ±√9 x = ±3

For u = -1: x^2 = -1 This has no real solutions because the square of a real number is always non-negative.

So, the solutions to the equation x^4 - 8x^2 - 9 = 0 are x = 3 and x = -3.

2. 16x^4 - 409x^2 + 225 = 0

This is also a quadratic equation in terms of x^2. Let u = x^2, then the equation becomes:

16u^2 - 409u + 225 = 0

Now, we can solve this quadratic equation for u. You can use the quadratic formula:

u = (-b ± √(b^2 - 4ac)) / (2a)

In this case, a = 16, b = -409, and c = 225. Plugging in these values:

u = (409 ± √((-409)^2 - 4 * 16 * 225)) / (2 * 16)

Calculate the discriminant:

D = (-409)^2 - 4 * 16 * 225 = 16641 - 14400 = 2241

Now, plug D back into the quadratic formula:

u = (409 ± √2241) / 32

u ≈ (409 ± 47.35) / 32

So, there are two possible values for u:

1. u ≈ (409 + 47.35) / 32 ≈ 14.16
2. u ≈ (409 - 47.35) / 32 ≈ 11.19

Now, we need to back-substitute to find the values of x:

For u ≈ 14.16: x^2 ≈ 14.16 x ≈ ±√14.16 x ≈ ±3.76

For u ≈ 11.19: x^2 ≈ 11.19 x ≈ ±√11.19 x ≈ ±3.35

So, the solutions to the equation 16x^4 - 409x^2 + 225 = 0 are x ≈ 3.76, x ≈ -3.76, x ≈ 3.35, and x ≈ -3.35.