Resolving Algebra Problem: Finding Real Solutions for \sqrt{2x-3} + x = 3

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In summary, the conversation is about a person struggling with an algebra problem and trying to solve it using different methods. They discuss the mistake in their first attempt and work through the problem using the completing the square method. They eventually reach the correct solution of x=2 for the original equation.
  • #1
sauri
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I got this algebra problem that I have been trying to get around but the answer I get is always different from the book.

It says: Find all the real solutions, if any of each equation.

21). [tex]\sqrt{2x-3} +x=3[/tex].

I tried to resolve it using the following method:

2x-3+x^2=9
(2x-12)+(x^2-9)=0
2(x-6)+(x+3)(x-3)=0
(x+3) and (x-3) will cancel each other so;
2x-12=0
x=12/2
x=6.

However, this is wrong and I know it, cause the answer does not fit the equation. So can anyone help me point what's wrong with what I did?
 
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  • #2
Yes, that's wrong! (a+ b)2 is not a2+ b2 so [itex](\sqrt{2x-3}+x)^2[/itex] is not 2x-3+ x!

I recommend you subtract x from both sides of th equation to get
[itex]\sqrt{2x-3}= 3- x[/itex], with only the square root on one side and then square both sides.
 
  • #3
If I use [itex]\sqrt{2x-3}= 3- x[/itex], I then worked it out to
2x-3=9-6x+x^2
2x+6x-x^2=9+3
8x-x^2-12=0
So I re-wrote as;
-x^2+8x-12=0
(x+6)=0 or (x+2)=0

but then the answers add up as x=-6 or x=-2. and that can't be right..
 
  • #4
You didn't quite factor it correctly.
 
  • #5
Try factoring -x^2+8x-12=0 again. You may want to move everything to the right hand side to make it a bit easier.
 
  • #6
O.k, this time I tried the copleting the square method:

-x^2+8x-12=0
-x^2+8x=12
(x+4)^2=12+16
[itex](x+4)=\sqrt28[/itex]
[itex]4+\sqrt28[/itex] or [itex]4-\sqrt28[/itex]
 
  • #7
The answers vary between -1.29 and 9.29, Still off
 
  • #8
-x^2 does not factor as (x+...)(x+...). That's your mistake.
 
  • #9
Rewrite -x2+ 8x- 12= 0 as x2- 8x+ 12= 0 by multiplying both sides by -1 (or "move everything to the right hand side" as was suggested before.

x2- 8x+ 12= (x- ?)(x- ?).

Of course, be sure to check solutions to the polynomial equation in the original equation!
 
  • #10
ah, I got it. You end up with (x- 6)(x- 2).
and only x=2 would apply for the equation.
Thanks
 

What does "Find all real solutions" mean?

"Find all real solutions" refers to finding all possible values for a given mathematical equation or problem that result in a real number as the solution. Real numbers are numbers that can be represented on a number line and include both positive and negative numbers, as well as zero.

Why is it important to find all real solutions?

Finding all real solutions is important because it provides a comprehensive understanding of the possible outcomes for a mathematical problem. It allows for a more accurate and complete analysis of the problem and can help identify any potential errors or inconsistencies in the given equation or problem.

How do you find all real solutions?

The process of finding all real solutions varies depending on the specific problem and equation. In general, it involves using algebraic techniques such as factoring, substitution, or the quadratic formula to solve the equation and determine all possible values for the variable. It is also important to check for any extraneous solutions, which may be values that satisfy the equation but are not actually valid.

Are there any cases where there are no real solutions?

Yes, there are cases where there are no real solutions. This can occur when the given equation or problem has imaginary solutions, which involve the use of complex numbers. Complex numbers include a real part and an imaginary part, and are not represented on a traditional number line. In these cases, the solutions may involve the use of the imaginary unit, denoted by "i".

Can there be more than one set of real solutions?

Yes, there can be more than one set of real solutions. This is most common in polynomial equations, where the degree of the equation determines the number of solutions. For example, a quadratic equation may have two real solutions, while a cubic equation may have three real solutions. It is important to carefully consider the degree of the equation and any possible restrictions on the variable when determining the number of real solutions.

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