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Helppp how to solve x=5√(x) - 6

x=5√(x) - 6


i know you take 6 to the other side then what


thankyou in advance

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Original post by monisj1
x=5√(x) - 6


i know you take 6 to the other side then what


thankyou in advance

Perhaps make the substitution y=xy=\sqrt{x}.
square both sides (you might need to check all solutions in your original equation as squaring can introduce extraneous solutions)
Original post by monisj1
x=5√(x) - 6


i know you take 6 to the other side then what


thankyou in advance


x−5x+6=0x - 5\sqrt{x} + 6 = 0

is a quadratic equation
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monisj1
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Original post by Arithmeticae
square both sides (you might need to check all solutions in your original equation as squaring can introduce extraneous solutions)



could you explain further with the question I'm zoo confused
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monisj1
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Original post by TenOfThem
x−5x+6=0x - 5\sqrt{x} + 6 = 0

is a quadratic equation


could you explain further with the question I'm zoo confused
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monisj1
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Original post by keromedic
Perhaps make the substitution y=xy=\sqrt{x}.


could you explain further with the question I'm zoo confused
Original post by monisj1
could you explain further with the question I'm zoo confused


Can you solve quadratic equations

If this said y2−5y+6=0y^2 - 5y + 6 = 0 could you solve it

Do you realise that x=(x)2x = (\sqrt{x})^2
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monisj1
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Original post by TenOfThem
Can you solve quadratic equations

If this said y2−5y+6=0y^2 - 5y + 6 = 0 could you solve it

Do you realise that x=(x)2x = (\sqrt{x})^2



so does y=x squared or square root of x
Original post by monisj1
so does y=x squared or square root of x


Look at your equation and mine

surely you can see what the y is replacing
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monisj1
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Original post by TenOfThem
Look at your equation and mine

surely you can see what the y is replacing



oh my god it just clicked in my head thank you very much
Original post by monisj1
oh my god it just clicked in my head thank you very much


good :biggrin:
So how do you work it out?
Original post by KittyRe-play
So how do you work it out?


Do you know how to solve quadratics?
Original post by KittyRe-play
So how do you work it out?

After getting y=somethingy=something, you know what x\sqrt{x} is == to.
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monisj1
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Original post by KittyRe-play
So how do you work it out?



basically my answer is x=9 and x=4
Original post by TenOfThem
Do you know how to solve quadratics?


Yes, but where is the y coming from?
Original post by monisj1
basically my answer is x=9 and x=4


How?
Original post by KittyRe-play
Yes, but where is the y coming from?


Nowhere

the equation is x−5x+6=0x - 5\sqrt{x} + 6 = 0

The equation with y in is just one that looks like it - with the same coefficients

If you can solve one, then you can solve the other
Original post by TenOfThem
Nowhere

the equation is x−5x+6=0x - 5\sqrt{x} + 6 = 0

The equation with y in is just one that looks like it - with the same coefficients

If you can solve one, then you can solve the other

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:

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