Question

Sqrt(x+5)+sqrt(x)=2x-15+2sqrt(x^2+5x)

Answer 1

Ответ: x=4.

Пошаговое объяснение:

√(x+5)+√x=2x-15+2*√(x²+5x)    ОДЗ: х+5≥0   х≥-5    x≥0   ⇒  х∈[0;+∞)

√(x+5)+√x=2x-15+2*√(x*(x+5))

√(x+5)+√x=2x-15+2*√x*√(x+5))

√(x+5)+√x=x+2*√x*√(x+5))+x+5-20

√(x+5)+√x=(√x)²+2*√x*√(x+5)+√(x+5)-20

√(x+5)+√x=(√x+√(x+5))²-20

(√x+√(x+5))²-(√(x+5)+√x)-20=0

Пусть √x+√(x+5)=t>0     ⇒

t²-t-20=0   D=81      √D=9

t₁=-4  ∉

t₂=5     ⇒

√x+√(x+5)=5

√(x+5)=5-√x

(√(x+5))²=(5-√x)²

x+5=25-2*5*√x+x

10*√x=20  |÷10

√x=2

(√x)²=2²

x=4 ∈ ОДЗ.

Answer 2

To solve the equation $\sqrt{x+5}+\sqrt{x}=2x-15+2\sqrt{x^2+5x}$, you can follow these steps:

1. Isolate the square roots on one side of the equation by moving $\sqrt{x}$ to the right side:

$\sqrt{x+5} = 2x - 15 + 2\sqrt{x^2+5x} - \sqrt{x}$

2. Square both sides of the equation to eliminate the square roots:

$(\sqrt{x+5})^2 = (2x - 15 + 2\sqrt{x^2+5x} - \sqrt{x})^2$

$x+5 = 4x^2 - 60x + 225 + 4(x^2 + 5x) - 4\sqrt{x(x^2+5x)}(2x-15) + (\sqrt{x})^2$

3. Simplify the equation:

$x+5 = 4x^2 - 60x + 225 + 4x^2 + 20x - 4\sqrt{x(x^2+5x)}(2x-15) + x$

4. Combine like terms:

$x+5 = 8x^2 - 40x + 225 - 4\sqrt{x(x^2+5x)}(2x-15)$

5. Move all terms to one side to set the equation equal to zero:

$8x^2 - 46x + 220 - 4\sqrt{x(x^2+5x)}(2x-15) - x - 5 = 0$

6. Combine like terms again:

$8x^2 - 47x + 215 - 4\sqrt{x(x^2+5x)}(2x-15) = 0$

7. Now, we have a quadratic equation in terms of $\sqrt{x(x^2+5x)}$. Let's isolate this term:

$4\sqrt{x(x^2+5x)}(2x-15) = 8x^2 - 47x + 215 - x - 5$

$4\sqrt{x(x^2+5x)}(2x-15) = 7x^2 - 46x + 210$

8. Divide both sides by $4(2x-15)$