Question

8x + (x-3)(x + 3) ≥ (x+4)²​

Answer 1

8x + (x-3)(x + 3) ≥ (x+4)²

х²+8х+16<=8х+х²-9

х²-х²+8х-8х+16+9<=0

25<=0

уравнение не имеет решений

Answer 2

Решение дано на фотографии.

Answer 3

Let's solve the given inequality step by step:

$8x + (x - 3)(x + 3) \geq (x + 4)^2$

First, let's expand and simplify each side of the inequality:

$8x + (x^2 - 9) \geq x^2 + 8x + 16$

Now, let's simplify further:

$8x + x^2 - 9 \geq x^2 + 8x + 16$

Next, let's rearrange the terms and isolate $x$ on one side of the inequality:

$x^2 - x^2 + 8x - 8x - 9 - 16 \geq 0$

$- 25 \geq 0$

Now, we can see that this inequality is not possible because $-25$ is not greater than or equal to $0$. Therefore, there are no solutions for $x$ that satisfy the given inequality $8x + (x - 3)(x + 3) \geq (x + 4)^2$.