SOLUTION: Find the exact and approximate solutions to {{{x^2-5*x+3=0}}}

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Question 130755: Find the exact and approximate solutions to x%5E2-5%2Ax%2B3=0
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Let's use the quadratic formula to solve for x:


Starting with the general quadratic

ax%5E2%2Bbx%2Bc=0

the general solution using the quadratic equation is:

x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29



So lets solve x%5E2-5%2Ax%2B3=0 ( notice a=1, b=-5, and c=3)




x+=+%28--5+%2B-+sqrt%28+%28-5%29%5E2-4%2A1%2A3+%29%29%2F%282%2A1%29 Plug in a=1, b=-5, and c=3



x+=+%285+%2B-+sqrt%28+%28-5%29%5E2-4%2A1%2A3+%29%29%2F%282%2A1%29 Negate -5 to get 5



x+=+%285+%2B-+sqrt%28+25-4%2A1%2A3+%29%29%2F%282%2A1%29 Square -5 to get 25 (note: remember when you square -5, you must square the negative as well. This is because %28-5%29%5E2=-5%2A-5=25.)



x+=+%285+%2B-+sqrt%28+25%2B-12+%29%29%2F%282%2A1%29 Multiply -4%2A3%2A1 to get -12



x+=+%285+%2B-+sqrt%28+13+%29%29%2F%282%2A1%29 Combine like terms in the radicand (everything under the square root)




x+=+%285+%2B-+sqrt%2813%29%29%2F2 Multiply 2 and 1 to get 2

So now the expression breaks down into two parts

x+=+%285+%2B+sqrt%2813%29%29%2F2 or x+=+%285+-+sqrt%2813%29%29%2F2



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So the exact solutions are


x+=+%285+%2B+sqrt%2813%29%29%2F2 or x+=+%285+-+sqrt%2813%29%29%2F2



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When you use a calculator, you will find that the approximate solutions are


x=4.30277563773199 or x=0.697224362268005