How Do You Solve Equations with Square Roots and Quadratic Terms?

[Stratosphere]

Homework Statement

A basic example is \sqrt{x-1}=x-7

Homework Equations

The Attempt at a Solution

I end up with an x^{2} and a normal x I don't know to to solve it.

 

[Integral]

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]

Homework Statement

A basic example is \sqrt{x-1}=x-7

Homework Equations

The Attempt at a Solution

I end up with an x^{2} and a normal x I don't 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.

 

[Stratosphere]

I thought this was a radical equation? Not a quadric one. If this is a quadric one I don't have to worry abought that yet then.
How do you solve somthing with a root on both side then? Like \sqrt{3x+2} + \sqrt{2x}= 35, I keep getting an answer that's too high.

 

[Office_Shredder]

Turn it into

\sqrt{3x+2} = 35 - \sqrt{2x}

square both sides

3x+2 = 35^2 + 2x - 70 \sqrt{2x}

and now you only have one square root

 

[symbolipoint]

Stratosphere, your first example, \sqrt{x-1} = 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.

 

[Stratosphere]

were does the -70 come from?

 

[General_Sax]

(35-\sqrt{2x})^2

Expand it out.

 

[Chaos2009]

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.

 

[Stratosphere]

Oh and how do you know what is a,b and c when using the quadratic formula?

 

[yeongil]

The general form of a quadratic equation is ax^{2} + bx + c = 0. Hopefully that answers your question. ;)


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