Question

(5x+1)^2≤(3x-2)^2+(4x+3)^2

Answer 1

(5х+1)²≤(3х-2)²+(4х+3)²
25х²+10х+1≤9х²-12х+4+16х²+24х+9
25х²+10х+1≤25х²+12х+13
10х-12х≤13-1
-2х≤12
х≥-6

Answer 2

To solve the inequality $(5x+1)^2 \leq (3x-2)^2 + (4x+3)^2$, we need to find the values of $x$ that satisfy the inequality. Let's proceed step by step:

Step 1: Expand both sides of the inequality:

$(5x+1)^2 = (5x+1)(5x+1) = 25x^2 + 10x + 1$

$(3x-2)^2 = (3x-2)(3x-2) = 9x^2 - 12x + 4$

$(4x+3)^2 = (4x+3)(4x+3) = 16x^2 + 24x + 9$

Step 2: Substitute these expanded expressions back into the original inequality:

$25x^2 + 10x + 1 \leq (9x^2 - 12x + 4) + (16x^2 + 24x + 9)$

Step 3: Simplify and combine like terms:

$25x^2 + 10x + 1 \leq 25x^2 + 12x + 13$

Step 4: Move all terms to one side of the inequality:

$0 \leq 25x^2 + 12x + 13 - (25x^2 + 10x + 1)$

Step 5: Simplify:

$0 \leq 25x^2 + 12x + 13 - 25x^2 - 10x - 1$

Step 6: Combine like terms again:

$0 \leq 2x + 12$

Step 7: Move the constant term to the other side of the inequality:

$-12 \leq 2x$

Step 8: Finally, divide by 2 (which doesn't change the direction of the inequality since 2 is positive):

$-6 \leq x$

So, the solution to the inequality is $x \geq -6$. In interval notation, the solution is $[-6, \infty)$.