Asked by Lydia
Can someone please help me with this one?
Solve (5x+4)^1/2-3x=0
Just so you know, it is to the 1/2 power.
Add 3x to both sides of the equation, then square both sides.
That should enable you to gather terms and factor.
Yeah, that's the part I understood. Could you help me with the factoring?
This is what I have as of now:
(5x+4)^1/2-3x=0
5x+4=3x^2
3x^2-5x-4=0
Go to the quadratic equation
x= -b +- sqrt (b^2=4ac) /2a
x= 5 +- sqrt (25+48) /6
x= 5 +- sqrt (73) /6
x= -0.590667291 ; x= 2.25733396
check my work.
Oh, really? We didn't learn about doing it like that, but I think it does work. Thanks!
wait, I got the same answers as you, but then I checked my work and neither worked
dang: error
(5x+4)^1/2-3x=0
5x+4=Mb>9</b>x^2
typo
(5x+4)^1/2-3x=0
5x+4= 9x^2
That changes the quadratic to
9x^2 -5x -4=0
x= ( 5 +- sqrt (25 + 36*4))/18
check that.
yes it works! thanks a ton!
Solve (5x+4)^1/2-3x=0
Just so you know, it is to the 1/2 power.
Add 3x to both sides of the equation, then square both sides.
That should enable you to gather terms and factor.
Yeah, that's the part I understood. Could you help me with the factoring?
This is what I have as of now:
(5x+4)^1/2-3x=0
5x+4=3x^2
3x^2-5x-4=0
Go to the quadratic equation
x= -b +- sqrt (b^2=4ac) /2a
x= 5 +- sqrt (25+48) /6
x= 5 +- sqrt (73) /6
x= -0.590667291 ; x= 2.25733396
check my work.
Oh, really? We didn't learn about doing it like that, but I think it does work. Thanks!
wait, I got the same answers as you, but then I checked my work and neither worked
dang: error
(5x+4)^1/2-3x=0
5x+4=Mb>9</b>x^2
typo
(5x+4)^1/2-3x=0
5x+4= 9x^2
That changes the quadratic to
9x^2 -5x -4=0
x= ( 5 +- sqrt (25 + 36*4))/18
check that.
yes it works! thanks a ton!
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