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Ivenika [448]
3 years ago
8

t="x^2(x-1)(x-4)-2x^3=-6x^2" align="absmiddle" class="latex-formula">
If x > 0, what is one possible solution to the equation above? (Hint: there are two real solutions)
Mathematics
2 answers:
RUDIKE [14]3 years ago
7 0

Answer:

x = 2 and 5

Step-by-step explanation:

x²(x - 1)(x - 4) - 2x³ = -6x² when x > 0

x²(x - 1)(x - 4) - 2x³ + 6x² = 0

~Use foil to simplify the stuff in parenthesis

x⁴ - 5x³ + 4x² - 2x³ + 6x² = 0

~Combine like terms

x⁴ - 7x³ + 10x² = 0

~Factor

x(x - 2) - 5(x - 2)

~Simplify

(x - 2)(x - 5) = 0

~Solve both factors

x - 2 = 0

x = 2

x - 5 = 0

x = 5

Best of Luck!

Sav [38]3 years ago
5 0

Answer:

  • 2 and 5

Step-by-step explanation:

<u>Solving in steps</u>

  • x^2(x - 1)(x - 4) - 2x^3 = -6x^2, x > 0
  • x^2(x^2 - 5x + 4) -2x^3 + 6x^2 = 0
  • x^4 - 5x^3 + 4x^2 - 2x^3 + 6x^2 = 0
  • x^4 - 7 x^3 + 10x^2 = 0
  • x^2 - 7x + 10 = 0
  • x^2 - 2x - 5x + 10 = 0
  • x(x - 2) - 5(x - 2) = 0
  • (x - 2)( x - 5) = 0
  1. x - 2 = 0 ⇒ x = 2
  2. x - 5 = 0 ⇒ x = 5
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