Question

X^2-7|x|+6=0(x-2)^2-8|x-2|+15=0x^2+|x+4|=4x^2+17=9x+4|x-3|

Answer 1

...............................................

Answer 2

To solve the given equations, we'll handle each equation separately:

1. $x^2 - 7|x| + 6 = 0$

Let's break this equation into two cases, one for $x \geq 0$ and another for $x < 0$.

For $x \geq 0$: $x^2 - 7x + 6 = 0$

Factoring: $(x - 1)(x - 6) = 0$

So, either $x - 1 = 0$ or $x - 6 = 0$. Solving for $x$: $x = 1$ or $x = 6$

For $x < 0$: $x^2 + 7x + 6 = 0$

Factoring: $(x + 1)(x + 6) = 0$

So, either $x + 1 = 0$ or $x + 6 = 0$. Solving for $x$: $x = -1$ or $x = -6$

Thus, the solutions for the equation are $x = -6, -1, 1, 6$.

2. $(x - 2)^2 - 8|x - 2| + 15 = 0$

Let's break this equation into two cases, one for $x \geq 2$ and another for $x < 2$.

For $x \geq 2$: $(x - 2)^2 - 8(x - 2) + 15 = 0$

Expanding: $x^2 - 4x + 4 - 8x + 16 + 15 = 0$ $x^2 - 12x + 35 = 0$

Factoring: $(x - 5)(x - 7) = 0$

So, either $x - 5 = 0$ or $x - 7 = 0$. Solving for $x$: $x = 5$ or $x = 7$

For $x < 2$: $(x - 2)^2 + 8(x - 2) + 15 = 0$

Expanding: $x^2 - 4x + 4 + 8x - 16 + 15 = 0$ $x^2 + 4x + 3 = 0$

Factoring: $(x + 1)(x + 3) = 0$

So, either $x + 1 = 0$ or $x + 3 = 0$. Solving for $x$: $x = -1$ or $x = -3$

Thus, the solutions for the equation are $x = -3, -1, 5, 7$.

3. $x^2 + |x + 4| = 4x^2 + 17 = 9x + 4|x - 3|$

This equation seems to be written incorrectly, as it contains two equal signs. Assuming you meant an equation, let's simplify it to find the solutions:

$x^2 + |x + 4| = 4x^2 + 17 - 9x - 4|x - 3|$

Let's break this equation into two cases, one for $x + 4 \geq 0$ and another for $x + 4 < 0$.

For $x + 4 \geq 0$: $x^2 + (x + 4) = 4x^2 + 17 - 9x - 4(x - 3)$