Question

Решите неравенство (2x+5)^2≥ (5x-2)^2

Answer 1

$(2x+5)^2\geq (5x-2)^2\\\\(2x+5)^2-(5x-2)^2\geq 0\; ,\qquad A^2-B^2=(A-B)(A+B)\\\\(2x+5-5x+2)(2x+5+5x-2)\geq 0\\\\(7-3x)(7x+3)\geq 0\\\\x_1=\frac{7}{3}\; ,\; \; x_2=-\frac{3}{7}\\\\znaki:\; \; \; +++[-\frac{3}{7}\, ]---[\, \frac{7}{3}\, ]+++\\\\x\in (-\infty ,-\frac{3}{7}\, ]\cup [\, 2\frac{1}{3}\; ,+\infty )$

Answer 2

To solve the inequality $(2x+5)^2 \geq (5x-2)^2$, we can use the property that for any real numbers $a$ and $b$, $a^2 \geq b^2$ if and only if $a \geq b$ or $a \leq -b$. In other words, the square of a real number is non-negative, so the inequality is preserved.

So, we have two cases to consider:

Case 1: $(2x+5)^2 \geq (5x-2)^2$ and $2x+5 \geq 5x-2$. Case 2: $(2x+5)^2 \geq (5x-2)^2$ and $2x+5 \leq - (5x-2)$.

Let's solve these two cases separately:

Case 1: $(2x+5)^2 \geq (5x-2)^2$

Expanding both sides: $4x^2 + 20x + 25 \geq 25x^2 - 20x + 4$

Bringing all terms to one side to form a quadratic inequality: $0 \geq 25x^2 - 4x^2 - 20x + 20x - 4 - 25$

Simplifying: $0 \geq 21x^2 - 29$

Now, we can solve for $x$ by setting $21x^2 - 29$ less than or equal to zero: $21x^2 - 29 \leq 0$

Solving for $x$: $21x^2 \leq 29$

Divide both sides by 21: $x^2 \leq \frac{29}{21}$

Now, take the square root of both sides (remembering to consider both positive and negative roots): $x \leq \pm \sqrt{\frac{29}{21}}$

Case 2: $(2x+5)^2 \geq (5x-2)^2$

Expanding both sides: $4x^2 + 20x + 25 \geq 25x^2 - 20x + 4$

Bringing all terms to one side to form a quadratic inequality: $0 \geq 25x^2 - 4x^2 - 20x + 20x - 4 - 25$

Simplifying: $0 \geq 21x^2 - 29$

This is the same inequality as in Case 1.

Therefore, the solutions to the inequality $(2x+5)^2 \geq (5x-2)^2$ are: $x \leq \sqrt{\frac{29}{21}}$ or $x \leq -\sqrt{\frac{29}{21}}$

Please note that in Case 2, there is no new solution. The inequality has the same solution set as in Case 1 because both sides of the original inequality are equal.