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Helppp how to solve x=5√(x) - 6

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Original post by KittyRe-play
So first I square the equation to get rid of the square root? That will give me x squared minus 25x plus 36? :s-smilie:


You can square both sides of the equation if you wish - that is an alternative method

but you have not done so correctly


Have you covered this topic in school?
Reply 21
Original post by KittyRe-play
How?

x5x+6=0x-5\sqrt{x}+6=0

Let y=xy=\sqrt{x} so that y2=xy^{2}=x

clearly this transforms our initial equation into y25y+6=0y^{2}-5y+6=0 which can then be solved for y, and then using the fact x=y2x=y^{2} we can find solutions for x.
Original post by TenOfThem
You can square both sides of the equation if you wish - that is an alternative method

but you have not done so correctly


Have you covered this topic in school?


Sorry i meant x squared minus 5x plus 36?? Do you not square the 5 as well?
Original post by KittyRe-play
Sorry i meant x squared minus 5x plus 36?? Do you not square the 5 as well?

You can't apply an operation to each element in an equation separately.
(a+b+c)2a2+b2+c2(a+b+c)^2 \neq a^2+b^2+c^2.
Eg.
36=(1+2+3)212+22+32=1436=(1+2+3)^2 \neq 1^2+2^2+3^2=14.

Two ways to solve x=5x6x=5\sqrt{x}-6 have been mentioned.
1)Treating it like a quadratic.
2)Squaring both sides.
(edited 9 years ago)
Original post by alpen
x5x+6=0x-5\sqrt{x}+6=0

Let y=xy=\sqrt{x} so that y2=xy^{2}=x

clearly this transforms our initial equation into y25y+6=0y^{2}-5y+6=0 which can then be solved for y, and then using the fact x=y2x=y^{2} we can find solutions for x.


Yeah so that would be y=3 & y=2 right?
Reply 25
Original post by keromedic
You can't apply an operation to each element in an equation separately.
(a+b+c)2a2+b2+c2(a+b+c)^2 \neq a^2+b^2+c^2.
Eg.
6=(1+2+3)212+22+32=146=(1+2+3)^2 \neq 1^2+2^2+3^2=14.

you just said 6=6^2
Original post by KittyRe-play
Sorry i meant x squared minus 5x plus 36?? Do you not square the 5 as well?


If you want to square then start with

x+6=5xx+6 = 5 \sqrt{x}

and square both sides



it is imo better to just solve the quadratic equation without squaring but you do not seem happy with the method shown in the thread - so try this
Original post by KittyRe-play
Yeah so that would be y=3 & y=2 right?


yes

so what is x
Reply 28
Original post by KittyRe-play
Yeah so that would be y=3 & y=2 right?

indeed, so x=3\sqrt{x}=3 or x=2\sqrt{x}=2
Original post by alpen
you just said 6=6^2

I edited it -_-.
Original post by alpen
indeed, so x=3\sqrt{x}=3 or x=2\sqrt{x}=2


Thank you. :smile:
Original post by TenOfThem
If you want to square then start with

x+6=5xx+6 = 5 \sqrt{x}

and square both sides



it is imo better to just solve the quadratic equation without squaring but you do not seem happy with the method shown in the thread - so try this


Yes indeed as I want to know this method. So if I square that do I get x^2+36=25x? What I don't get is when squaring do you square the coefficient of the square root of x, which in this case is 5?
Original post by KittyRe-play
Yes indeed as I want to know this method. So if I square that do I get x^2+36=25x? What I don't get is when squaring do you square the coefficient of the square root of x, which in this case is 5?


the 25x is fine but the (x+6)2(x+6)^2 does not equal x2+36x^2 + 36
I get it guys thanks a lot I guess that y method is better than what I was going for! Cheers. :smile:
Original post by TenOfThem
the 25x is fine but the (x+6)2(x+6)^2 does not equal x2+36x^2 + 36


Its x^2+12x+36=25x?
Original post by KittyRe-play
I get it guys thanks a lot I guess that y method is better than what I was going for! Cheers. :smile:


ok :smile:

it is the preferred method anyway :biggrin:
Original post by KittyRe-play
Its x^2+12x+36=25x?


yes, it is
Original post by TenOfThem
yes, it is


So then I minus 25x which gives x^2-13x+36=0
Solving this quadratic then gives x=4 and x=9
Hallaluya I get it! :biggrin:
Original post by KittyRe-play
So then I minus 25x which gives x^2-13x+36=0
Solving this quadratic then gives x=4 and x=9
Hallaluya I get it! :biggrin:


:biggrin:
Original post by TenOfThem
:biggrin:


But then using the y method the solutions are x=3 & x=2 and it's not the same as 4 and 9. So is that incorrect?

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