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3 years ago

3x²+x=2solve by factoring ?

2 answers:

8 0

$3x^2+x=2\\\\3x^2+x-2=0\\\\3x^2+3x-2x-2=0\\\\3x(x+1)-2(x+1)=0\\\\(x+1)(3x-2)=0\\\\x+1=0\ \ \ \ \ \ \ \ \ or\ \ \ \ \ \ \ \ \ 3x-2=0\\x=-1\ \ \ \ \ \ \ \ \ \ \ \ or\ \ \ \ \ \ \ \ \ 3x=2\\x=-1\ \ \ \ \ \ \ \ \ \ \ \ or\ \ \ \ \ \ \ \ \ x=\frac{2}{3}\\\\$

kondor19780726 [428]3 years ago

7 0

$3x^2+x=2\\
3x^2+x-2=0\\
3x^2+3x-2x-2=0\\
3x(x+1)-2(x+1)=0\\
(3x-2)(x+1)=0\\
x=\frac{2}{3} \vee x=-1$

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$\frac{-m^2-6m+9}{(m+3)(3-m)}=\frac{m^2+9}{(m+3)(m-3)}$

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- Multiply both sides by $m-3$ :

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- Apply Distributive property and simplify:

$(-m^2-6m+9)(m-3)=(m^2+9)(3-m)\\\\-m^3-6m^2+9m+3m^2+18m-27=3m^2+27-m^3-9m\\\\-m^3-3m^2+27m-27+m^3-3m^2+9m-27=0\\\\-6m^2+36m-54=0$

- Divide both sides of the equation by -6:

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- Factor the equation and solve for "m":

$(m-3)^2=0\\\\m=3$

In order to verify it, you must substitute $m=3$ into the equation and solve it:

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