SOLUTION: Can I get help please! It asks to solve by whatever method seems easiest to you. Leave answers in simplest radical form. The problem is 3x^2=2x+5 Please help me Thank You=)

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Question 131486This question is from textbook structure and method
: Can I get help please! It asks to solve by whatever method seems easiest to you. Leave answers in simplest radical form. The problem is 3x^2=2x+5
Please help me Thank You=) This question is from textbook structure and method

Found 3 solutions by 067isa, ijeoma, solver91311:
Answer by 067isa(13) About Me  (Show Source):
You can put this solution on YOUR website!
3x^2=2x+5
so zero the equation by moving all the terms in one side
3x^2-2x-5=0
factor this now
(AC method)= 5*3= 15
two number the prdouct of them will be 15 and the differece between them is 2
which will be 3 and 5
now do the grouping
3x^2+3X-5x-5=0
3x(x+1)-5(x+1)=(x+1)(3x-5)=0
so now the two value that satisified the equation is x= -1 and 5/3

Answer by ijeoma(3) About Me  (Show Source):
You can put this solution on YOUR website!
3x^2-2x-5=0
factor out
3x-5=-15
look for two numbers that when u multiply u will get -15 and when u add both u will get -2
that number is 3 and -5
so, (x+3) and (x-5)
so, wat about the the coefficient of x^2=3
so we multiply both factors with 1/3
x=-3(i/3) and x=5(1/3)
therefore x=1- and x=5/3
u can as well use the quadratic formula and u will get same answer. hope this helps

Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!
3x%5E2=2x%2B5

The first step is to put the equation into standard form (ax%5E2%2Bbx%2Bc=0, so:

Add -2x-5 to both sides:

3x%5E2-2x-5=0

There are three methods to solve this. The first is by factoring the trinomial into to binomials:

%283x-5%29%28x%2B1%29=0, therefore 3x-5=0 or x%2B1=0, and x=5%2F3 or x=-1

The second method is to Complete the Square:

1. Put the constant term back on the right:
3x%5E2-2x=5

2. Divide through by the lead coefficient:
x%5E2-2x%2F3=5%2F3

3. Divide the coefficient on the x term by 2, square it, and add the result to both sides. %28-2%2F3%29%2F2=-1%2F3, %28-1%2F3%29%5E2=1%2F9, so:
x%5E2-2x%2F3%2B1%2F9=5%2F3%2B1%2F9

4. Simplify the right:
x%5E2-2x%2F3%2B1%2F9=16%2F9

5. Now that the left is a perfect square, factor it:
%28x-%281%2F3%29%29%5E2=16%2F9

6. Take the square root of both sides:
x-%281%2F3%29=sqrt%2816%2F9%29 or x-%281%2F3%29=-sqrt%2816%2F9%29
x-%281%2F3%29=4%2F3 or x-%281%2F3%29=-4%2F3
x=5%2F3 or x=-3%2F3=-1

The third way is to use the quadratic formula:
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+. For your problem, a = 3, b = -2, and c = -5, so:

x+=+%28-%28-2%29+%2B-+sqrt%28+%28-2%29%5E2-4%2A3%2A%28-5%29%29%29%2F%282%2A3%29+
x+=+%282+%2B-+sqrt%28+4-%28-60%29%29%29%2F6+
x+=+%282+%2B-+sqrt%2864%29%29%2F6+
x+=+%282+%2B+8%29%2F6+ or x+=+%282+-+8%29%2F6+
x+=+10%2F6+=5%2F3 or x+=+-6%2F6+=-1

Which way seems easiest to you? You tell me. The factoring method certainly looks easy, but in truth, I didn't see the factorization until after I had used the quadratic formula and found that there were rational roots to the equation. So the quadratic formula was really the easiest for me. I generally avoid completing the square when dividing by the lead coefficient or dividing the first degree term coefficient by 2 results in a fraction because it can become a messy calculation. The fact that this problem had rational roots made it all come out very neatly, but neat perfect squares are not something that you will generally find in real life applications.