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MakcuM [25]
2 years ago
12

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

Mathematics
2 answers:
Montano1993 [528]2 years ago
6 0

x^2(4x-3)(5x-1)=0\iff x^2=0\ \vee\ 4x-3=0\ \vee\ 5x-1=0\\\\x^2=0\to\boxed{x=0}\\\\4x-3=0\ \ \ |+3\\4x=3\ \ \ |:4\\\boxed{x=\dfrac{3}{4}}\\\\5x-1=0\ \ \ |+1\\5x=1\ \ \ |:5\\\boxed{x=\dfrac{1}{5}}\\\\Answer:\ x=0\ or\ x=\dfrac{3}{4}\ or\ x=\dfrac{1}{5}

kati45 [8]2 years ago
3 0

Hi!

x = 0

or

x = 1/5

or

x = 3/4

---

The process:

Step 1: Factor left side of equation.

Step 2: Set factors equal to 0.

-

Hope this helped :)

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