solving radical equations with one variable
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sqrt(3a+10)=sqrt(2a-1)+2

To remove the radical on the left-hand side of the equation, square both sides of the equation.
(~(3a+10))^(2)=(~(2a-1)+2)^(2)

Simplify the left-hand side of the equation.
3a+10=(~(2a-1)+2)^(2)

Squaring an expression is the same as multiplying the expression by itself 2 times.
3a+10=(~(2a-1)+2)(~(2a-1)+2)

Multiply each term in the first group by each term in the second group using the FOIL method. FOIL stands for First Outer Inner Last, and is a method of multiplying two binomials.  First, multiply the first two terms in each binomial group.  Next, multiply the outer terms in each group, followed by the inner terms.  Finally, multiply the last two terms in each group.
3a+10=(~(2a-1)*~(2a-1)+~(2a-1)*2+2*~(2a-1)+2*2)

Simplify the FOIL expression by multiplying and combining all like terms.
3a+10=(~(2a-1)^(2)+4~(2a-1)+4)

Remove the parentheses around the expression ~(2a-1)^(2)+4~(2a-1)+4.
3a+10=~(2a-1)^(2)+4~(2a-1)+4

Raising a square root to the square power results in the expression inside the root.
3a+10=(2a-1)+4~(2a-1)+4

Add 4 to -1 to get 3.
3a+10=2a+3+4~(2a-1)

Since a is on the right-hand side of the equation, switch the sides so it is on the left-hand side of the equation.
2a+3+4~(2a-1)=3a+10

Move all terms not containing ~(2a-1) to the right-hand side of the equation.
4~(2a-1)=-2a-3+3a+10

Simplify the right-hand side of the equation.
4~(2a-1)=a+7

Divide each term in the equation by 4.
(4~(2a-1))/(4)=(a)/(4)+(7)/(4)

Simplify the left-hand side of the equation by canceling the common terms.
~(2a-1)=(a)/(4)+(7)/(4)

To remove the radical on the left-hand side of the equation, square both sides of the equation.
(~(2a-1))^(2)=((a)/(4)+(7)/(4))^(2)

Simplify the left-hand side of the equation.
2a-1=((a)/(4)+(7)/(4))^(2)

Combine the numerators of all expressions that have common denominators.
2a-1=((a+7)/(4))^(2)

Expand the exponent of 2 to the inside factor (a+7).
2a-1=((a+7)^(2))/((4)^(2))

Expand the exponent 2 to 4.
2a-1=((a+7)^(2))/(4^(2))

Simplify the exponents of 4^(2).
2a-1=((a+7)^(2))/(16)

Multiply each term in the equation by 16.
2a*16-1*16=((a+7)^(2))/(16)*16

Simplify the left-hand side of the equation by multiplying out all the terms.
32a-16=((a+7)^(2))/(16)*16

Simplify the right-hand side of the equation by simplifying each term.
32a-16=(a+7)^(2)

Since (a+7)^(2) contains the variable to solve for, move it to the left-hand side of the equation by subtracting (a+7)^(2) from both sides.
32a-16-(a+7)^(2)=0

Squaring an expression is the same as multiplying the expression by itself 2 times.
32a-16-((a+7)(a+7))=0

Multiply -1 by each term inside the parentheses.
32a-16-a^(2)-14a-49=0

Since 32a and -14a are like terms, add -14a to 32a to get 18a.
18a-16-a^(2)-49=0

Subtract 49 from -16 to get -65.
18a-65-a^(2)=0

Move all terms not containing a to the right-hand side of the equation.
-a^(2)+18a-65=0

Multiply each term in the equation by -1.
a^(2)-18a+65=0

For a polynomial of the form x^(2)+bx+c, find two factors of c (65) that add up to b (-18).  In this problem -5*-13=65 and -5-13=-18, so insert -5 as the right hand term of one factor and -13 as the right-hand term of the other factor.
(a-5)(a-13)=0

Set each of the factors of the left-hand side of the equation equal to 0.
a-5=0_a-13=0

Since -5 does not contain the variable to solve for, move it to the right-hand side of the equation by adding 5 to both sides.
a=5_a-13=0

Set each of the factors of the left-hand side of the equation equal to 0.
a=5_a-13=0

Since -13 does not contain the variable to solve for, move it to the right-hand side of the equation by adding 13 to both sides.
a=5_a=13

The complete solution is the set of the individual solutions.
a=5,13
by Level 6 User (23.6k points)

The first thing to realise is that, since we can only have a real square root of a positive number, then 3a+10 and 2a-1 must be positive so a must be greater than -10/3 and greater than 1/2. That just means of course that a is bigger than 1/2, since 1/2 is already bigger than -10/3.  Now, square both sides of the equation.

3a+10=2a-1+4+4sqrt(2a-1)

a+7=4sqrt(2a-1)

Square both sides again.

a^2+14a+49=32a-16 which becomes the quadratic equation a^2-18a+65=0. This factorises into (a-5)(a-13)=0, from which a=5 and 13, both of which satisfy the requirements for keeping the square root expressions positive. Substituting these values for a in the original question we can see that they both fit.

by Top Rated User (1.1m points)

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