Question

(2х+2)^2>(х-5)^2Даю 20 баллов ​

Answer 1

Ответ:

(2x+2)^2>(x-5)^2

2x+2>x-5

x>-7

Answer 2

To solve the inequality $(2x+2)^2 > (x-5)^2$, we need to find the values of $x$ that satisfy the inequality. Let's break down the steps to solve it:

Step 1: Expand both sides of the inequality: $(2x+2)^2 = (2x+2)(2x+2) = 4x^2 + 8x + 4$ $(x-5)^2 = (x-5)(x-5) = x^2 - 10x + 25$

Step 2: Substitute the expanded expressions back into the inequality: $4x^2 + 8x + 4 > x^2 - 10x + 25$

Step 3: Move all terms to one side of the inequality to set it to zero: $4x^2 + 8x + 4 - x^2 + 10x - 25 > 0$

Step 4: Simplify the inequality: $3x^2 + 18x - 21 > 0$

Step 5: Factor the quadratic expression, if possible. In this case, we can't factor it easily, so we can use the quadratic formula:

The quadratic formula is given by $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ for the quadratic equation $ax^2 + bx + c = 0$.

For our inequality $3x^2 + 18x - 21 > 0$, we have $a = 3$, $b = 18$, and $c = -21$.

Step 6: Apply the quadratic formula to find the critical points (where the expression equals zero): $x = \frac{-18 \pm \sqrt{18^2 - 4 \cdot 3 \cdot (-21)}}{2 \cdot 3}$

Solving this equation, we get two critical points: $x = \frac{-18 + \sqrt{468}}{6} \approx 1.225$ and $x = \frac{-18 - \sqrt{468}}{6} \approx -6.225$.

Step 7: Determine the sign of the expression in the intervals formed by the critical points.

Test a value in each interval:

• For $x < -6.225$, let's use $x = -7$: $3(-7)^2 + 18(-7) - 21 = 147 > 0$

• For $-6.225 < x < 1.225$, let's use $x = 0$: $3(0)^2 + 18(0) - 21 = -21 < 0$

• For $x > 1.225$, let's use $x = 2$: $3(2)^2 + 18(2) - 21 = 27 > 0$

Step 8: Analyze the results:

• The expression is positive for $x < -6.225$ and $x > 1.225$.
• The expression is negative for $-6.225 < x < 1.225$.

Step 9: Final conclusion:

The inequality $(2x+2)^2 > (x-5)^2$ is satisfied for $x < -6.225$ and $x > 1.225$.

So, the solution to the inequality is $-\infty < x < -6.225$ or $x > 1.225$.