7. Solve for x and check

 

3x^2-x-5=0

 

This will not factor, so the simplest method for solving is to use

the quadratic formula

 

x=(-b+-Sqrt(b^2-4ac))/(2a)

 

where

 

Substitute these numbers into the formula.

 

 

x=(-(-1)+-Sqrt((-1)^2-4(3)(-5))/(2(3)

 

Do the arithmetic under the radical: first the power, then the

product.

 

x=(-(-1)+-Sqrt(1+60))/(2(3))

 

When we finish the arithmetic under the radical, we may as

well remove the parentheses outside the radical.

 

 

x=(1+-Sqrt61)/6

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Check:

 

The checks to these equations are extremely fascinating. Let us

first check the solution

 

x=(1+Sqrt61)/6

 

Copy down the original equation

 

3x^2-x-5=0

 

except that wherever you see an x, copy down the solution in

parentheses.

 

 

3((1+Sqrt61)/6)^2- ((1+Sqrt61)/6)-5=0

 

 

 

This gives us an extensive order of operations problem. Since

the expressions inside the parentheses are as simplified as possible,

we do the powers first. There is a square in the first term. Note that

the factor being squared is a fraction with a binomial on the top.

When we square the top and square the bottom, we use the

technique for squaring binomials.

 

(a+b)^2=a^2+2ab+b^2

 

This gives us

 

3((1+2Sqrt61+61)/36)- ((1+Sqrt61)/6)-5=0

 

 

When we cancel the 3 and the 26, the first fraction has a

denominator of 12, so we would need to multiply the top and bottom

of the second fraction by 2 to get common denominators.

 

 

(1+2Sqrt61+61)/12+2(1+Sqrt61)/(2(6)-5=0

 

 

 

We could easily turn the 5 into a 60/12 to get it over a

common denominator as well, but one finds that when one does

these checks that if we simplify the fractions we already have at this

point, it will make our lives simpler. Note how the radical terms

vanish leaving us with

 

(1+2Sqrt61+61-2-2Sqrt61)/12-5=0

 

The fraction simplifies to

 

(1+61-2)/12-5=0

 

 

60/12-5=0

or

 

5-5=0

 

which checks.

 

The check of

 

x=(1-Sqrt61)/6

 

is entirely similar and is left to the reader.

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