Simple Equation: Find the Roots of 2x+3√(3x-5)=5 using Algebraic Manipulation

[Maskotti]

3.4 10

$$2x+3\sqrt{3x-5}=5$$

$$3\sqrt{3x-5}=5-2x \ \ \ \ \ \ \mid ()^2$$

$$\\9(3x-5)=25+2*5*(-2x)+(2x)^2$$

$$\\27x-45=25-20x+4x^2$$

$$\\27x+20x-4x^2=25+45$$

$$\\47x-4x^2=70$$

$$\\-4x^2+47x-70=0$$

hmhph

$$x= \frac { -47^+_- \sqrt {47^2-4*(-4)*-(70)}}{2*-4}$$


$$x= \frac { -47^+_-33}{-8}$$

$$x= \ 10 \ or \ x= \ \frac {7}{4}$$

 

[recon]

It seems right.

What exactly is your question here?

 

[HallsofIvy]

I hope you realize that writing
$$\\x \left( x - \frac {47}{4} \right) = \frac{70}{4}$$
doesn't help you at all: if ab=0 then either a=0 or b= 0 but that only works for "= 0".

You could have written $4x^2- 47x-70=0$ and perhaps have reconized that this can be factored: (x-10)(4x-7) but I will confess that I got that by looking at your solution!