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Solving For X on both sides

1. Homework Statement
A basic example is [tex]\sqrt{x-1}[/tex]=x-7


2. Homework Equations



3. The Attempt at a Solution
I end up with an x[tex]^{2}[/tex] and a normal x I dont know to to solve it.
 

Integral

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Please show us your work. You say you have and x2. ok, you should have an x2 term, now show us exactly what you have and how you got it.
 

HallsofIvy

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1. Homework Statement
A basic example is [tex]\sqrt{x-1}[/tex]=x-7


2. Homework Equations



3. The Attempt at a Solution
I end up with an x[tex]^{2}[/tex] and a normal x I dont know to to solve it.
So that's a quadratic equation, isn't it? You can solve quadratic equations by factoring, completing the square, or using the quadratic formula.
 
I thought this was a radical equation? Not a quadric one. If this is a quadric one I dont have to worry abought that yet then.
How do you solve somthing with a root on both side then? Like [tex]\sqrt{3x+2}[/tex] + [tex]\sqrt{2x}[/tex]= 35, I keep getting an answer thats too high.
 
Last edited:

Office_Shredder

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Turn it into

[tex] \sqrt{3x+2} = 35 - \sqrt{2x}[/tex]

square both sides

[tex]3x+2 = 35^2 + 2x - 70 \sqrt{2x}[/tex]

and now you only have one square root
 

symbolipoint

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Stratosphere, your first example, [tex]\sqrt{x-1}[/tex] = x - 7, should give you no complications when you square both sides. It WILL give you a quadratic equation. Either you can (after suitable algebraic steps) factor and solve, or you can complete the square/use solution to quadratic equation to solve.
 
were does the -70 come from?
 
[tex] (35-\sqrt{2x})^2 [/tex]

Expand it out.
 
Don't forget, when you square the radical, you can end up finding solutions for x that do not work. You have to go through and check these solutions with your original equation to make sure that each value works.
 
Oh and how do you know what is a,b and c when using the quadratic formula?
 
The general form of a quadratic equation is [tex]ax^{2} + bx + c = 0[/tex]. Hopefully that answers your question. ;)


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