Question

(8x-3)²- 10|8x-3| -24=0 (x² +2x+2) (x²+2x-4)=-5

Answer 1

$(8x-3)^{2} -10|8x-3|-24=0\\\\|8x-3|^{2}-10|8x-3|-24=0\\\\\left[\begin{array}{ccc}|8x-3|=12\\|8x-3|=-2-\oslash\end{array}\right \\\\|8x-3|=12\\\\\left[\begin{array}{ccc}8x-3=12\\8x-3=-12\end{array}\right\\\\\\\left[\begin{array}{ccc}x=1,875\\x=-1,125\end{array}\right$

(x² + 2x + 2)(x² + 2x - 4) = - 5

Сделаем замену :

x² + 2x - 4 = m   ⇒  x² + 2x + 2 = m + 6

(m + 6) * m + 5 = 0

m² + 6m + 5 = 0

По теореме Виета :

m₁ = - 1        m₂ = - 5

1)  x² + 2x - 4 = - 1

x² + 2x - 3 = 0

x₁ = - 3       x₂ = 1

2) x² + 2x - 4 = - 5

x² + 2x + 1 = 0

(x + 1)² = 0

x + 1 = 0

x₃ = -1

Ответ : - 3 ; -1 ; 1

Answer 2

$(8x-3)^2-10|8x-3|-24=0\Leftrightarrow |8x-3|^2-10|8x-3|-24=0\\D_{|8x-3|}=25+24=49\\|8x-3|=5\pm 7\Leftrightarrow |8x-3|=12\Leftrightarrow (8x-3-12)(8x-3+12)=0\Leftrightarrow \\\Leftrightarrow (8x-15)(8x+9)=0\Rightarrow x=\begin{Bmatrix}-\frac{9}{8};\frac{15}{8}\end{Bmatrix}$

$(x^2+2x+2)(x^2+2x-4)=-5\Leftrightarrow (x^2+2x)^2-2(x^2+2x)-3=0\\ax^2+bx+c=0\\a-b+c=0\Rightarrow x=\begin{Bmatrix}-1;-\frac{c}{a}\end{Bmatrix}\\1+2-3=0\Rightarrow x^2+2x=\begin{Bmatrix}-1;3\end{Bmatrix}\\x^2+2x+1=0\Leftrightarrow (x+1)^2=0\Rightarrow x=-1\\x^2+2x-3=0\Rightarrow x=\begin{Bmatrix}1;-3\end{Bmatrix}\\x=\begin{Bmatrix}-3;-1;1\end{Bmatrix}$

Answer 3

Let's solve each equation separately:

1. Solving (8x - 3)² - 10|8x - 3| - 24 = 0:

Let's consider two cases for the absolute value term:

Case 1: (8x - 3) ≥ 0 In this case, the absolute value term |8x - 3| is equal to (8x - 3). Substituting it in the equation:

(8x - 3)² - 10(8x - 3) - 24 = 0 Expanding and simplifying:

64x² - 48x + 9 - 80x + 30 - 24 = 0 64x² - 128x + 15 = 0

Case 2: (8x - 3) < 0 In this case, the absolute value term |8x - 3| is equal to -(8x - 3), which simplifies to (-8x + 3). Substituting it in the equation:

(8x - 3)² - 10(-8x + 3) - 24 = 0 Expanding and simplifying:

64x² - 48x + 9 + 80x - 30 - 24 = 0 64x² + 32x - 45 = 0

Now we have two quadratic equations to solve.

Case 1: Solving 64x² - 128x + 15 = 0:

Using the quadratic formula, x = (-b ± √(b² - 4ac)) / (2a), where a = 64, b = -128, and c = 15.

x = (-(-128) ± √((-128)² - 4(64)(15))) / (2(64)) x = (128 ± √(16384 - 3840)) / 128 x = (128 ± √12544) / 128 x = (128 ± 112) / 128

So, x₁ = (128 + 112) / 128 = 240 / 128 = 15/8 And x₂ = (128 - 112) / 128 = 16 / 128 = 1/8

Case 2: Solving 64x² + 32x - 45 = 0:

Using the quadratic formula, x = (-b ± √(b² - 4ac)) / (2a), where a = 64, b = 32, and c = -45.

x = (-(32) ± √((32)² - 4(64)(-45))) / (2(64)) x = (-32 ± √(1024 + 11520)) / 128 x = (-32 ± √(12544)) / 128 x = (-32 ± 112) / 128

So, x₃ = (-32 + 112) / 128 = 80 / 128 = 5/8 And x₄ = (-32 - 112) / 128 = -144 / 128 = -9/8

Therefore, the solutions to the equation (8x - 3)² - 10|8x - 3| - 24 = 0 are x = 15/8, 1/8, 5/8, and -9/8.

2. Solving (x² + 2x + 2)(x² + 2x - 4) = -5:

Expanding the left side of the equation:

x⁴ + 2x³ - 4x² + 2x³ + 4x² - 8x + 2x² + 4x - 8 = -5 Combining like terms:

x⁴ + 4x³ + 2x² - 4x - 13 = 0

To solve this quartic equation, a general formula does exist, but it is quite complex. We can approximate solutions numerically or by using computational methods. However, there is no straightforward algebraic solution.

If you need a numerical approximation for the roots of this equation, please let me know and provide a desired level of precision.