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Anvisha [2.4K]
3 years ago
11

How many integer solutions are there to the equation x^3-5x^2+2x+13=5

Mathematics
2 answers:
LuckyWell [14K]3 years ago
7 0
Hello,

To find the solutions to this, you can either graph it on a graphing calculator or factor it.

If you factor it, first set it equal to 0 by subtracting 5 from both this.
The you can factor it to: (x - 2)(x - 4)(x + 1) = 0
Using the zero product rule, our solutions are 2, 4, and -1.

There are 3 integer solutions.

Good luck,
MrEQ
vagabundo [1.1K]3 years ago
6 0
Three possible solutions
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Step-by-step explanation:

Given fraction is:

\dfrac{3}{m+3} -\dfrac{m}{3-m} = \dfrac{m^2+9}{m^2-9}\\\Rightarrow \dfrac{3}{m+3} +\dfrac{m}{m-3} = \dfrac{m^2+9}{m^2-9}\\\Rightarrow \dfrac{3\times (m-3)+ m(m+3)}{(m+3)(m-3)}  = \dfrac{m^2+9}{m^2-9}\\\Rightarrow \dfrac{3 m-9+ m^2+3m}{m^2-9}  = \dfrac{m^2+9}{m^2-9}\\\Rightarrow \dfrac{m^2-9+ 6m}{m^2-9}  = \dfrac{m^2+9}{m^2-9}\\If\ m^2-9 \neq 0\ or\ m \neq 3\\\Rightarrow m^2-9+ 6m  = m^2+9\\\Rightarrow 6m = 18\\\Rightarrow m = 3

<u>Formula used:</u>

<u></u>(a+b)(a-b) = a^{2} -b^{2}<u></u>

<u></u>

But, the above equation was solved at the condition that:

m\neq 3

So, there is no solution for the given equation.

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