Question

(x+5x)²-4(x²+5x)-12=0

Answer 1

$\displaystyle \tt (x+5x)^2-4(x^2+5x)-12=0\\\displaystyle \tt (6x)^2-4x^2-20x-12=0\\\displaystyle \tt 36x^2-4x^2-20x-12=0\\\displaystyle \tt 32x^2-20x-12=0 \: \: \: \: | \div4\\\displaystyle \tt 8x^2-5x-3=0\\\displaystyle \tt D=(-5)^2-4\cdot8\cdot(-3)=25+96=121\\\displaystyle \tt \sqrt{D}=\sqrt{121}=11\\\displaystyle \tt x_1=\frac{5+11}{2\cdot8}=\frac{16}{16}=1\\\displaystyle \tt x_2=\frac{5-11}{2\cdot8}=\frac{-6}{16}=-\frac{3}{8}=-0,375$

Answer 2

To solve the quadratic equation (x+5x)² - 4(x²+5x) - 12 = 0, let's simplify the equation first:

We can simplify the equation step by step:

(x+5x)² = (6x)² = 36x²

-4(x²+5x) = -4x² - 20x

Now, substituting these simplified terms into the equation, we get:

36x² - 4x² - 20x - 12 = 0

Combine like terms:

32x² - 20x - 12 = 0

Now, we have a quadratic equation in standard form, ax² + bx + c = 0, where:

a = 32 b = -20 c = -12

To solve this quadratic equation, we can use the quadratic formula:

x = (-b ± √(b² - 4ac)) / (2a)

Plugging in the values, we have:

x = (-(-20) ± √((-20)² - 4 * 32 * (-12))) / (2 * 32)

Simplifying further:

x = (20 ± √(400 + 1536)) / 64

x = (20 ± √1936) / 64

x = (20 ± 44) / 64

Now, we have two possible solutions:

x₁ = (20 + 44) / 64 = 64 / 64 = 1

x₂ = (20 - 44) / 64 = -24 / 64 = -3/8

Therefore, the solutions to the quadratic equation (x+5x)² - 4(x²+5x) - 12 = 0 are x = 1 and x = -3/8.