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# x is a real number, x*sqrt(x)-10*sqrt(x)-3=0; x+1/x=?

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x is a real number, x*sqrt(x)-10*sqrt(x)-3=0; x+1/x=? Feb 8, 2016

#6
+15

Let  sqrt(x) = y, and the first equation becomes

$$\displaystyle y^{3} - 10y - 3 = 0.$$

This clearly has a solution y = -3, so removing the factor (y + 3), the equation can be written

$$\displaystyle (y+3)(y^{2}-3y-1)=0,$$

yielding two further solutions for y,

$$\displaystyle y = (3 \pm\sqrt{13})/2.$$

That suggests three possible values for sqrt(x), but, (and maybe it's implied by the question ?), surely it is reasonable to assume that sqrt(x) is positive ?

If that's the case, then

$$\displaystyle \sqrt{x}=(3+\sqrt{13})/2.$$

Squaring that,

$$\displaystyle x = (11 + 3\sqrt{13})/2$$, so that

$$\displaystyle x + \frac{1}{x}=\frac{11+3\sqrt{13}}{2}+\frac{2}{11+3\sqrt{13}}$$,

$$\displaystyle =\frac{(11+3\sqrt{13})^{2}+2^{2}}{2(11+3\sqrt{13})}=\frac{121+33\sqrt{13}}{11+3\sqrt{13}} = 11.$$

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Feb 18, 2016

#1
+10

Solve for x: -3-10 sqrt(x)+x^(3/2) = 0      Simplify and substitute y = sqrt(x): -3-10 sqrt(x)+x^(3/2) = -3-10 sqrt(x)+sqrt(x)^3 = y^3-10 y-3 = 0: y^3-10 y-3 = 0 The left hand side factors into a product with two terms: (y+3) (y^2-3 y-1) = 0 Split into two equations: y+3 = 0 or y^2-3 y-1 = 0 Subtract 3 from both sides: y = -3 or y^2-3 y-1 = 0 Substitute back for y = sqrt(x): sqrt(x) = -3 or y^2-3 y-1 = 0 Raise both sides to the power of two: x = 9 or y^2-3 y-1 = 0 Add 1 to both sides: x = 9 or y^2-3 y = 1 Add 9/4 to both sides: x = 9 or y^2-3 y+9/4 = 13/4 Write the left hand side as a square: x = 9 or (y-3/2)^2 = 13/4 Take the square root of both sides: x = 9 or y-3/2 = sqrt(13)/2 or y-3/2 = -sqrt(13)/2 Add 3/2 to both sides: x = 9 or y = 3/2+sqrt(13)/2 or y-3/2 = -sqrt(13)/2 Substitute back for y = sqrt(x): x = 9 or sqrt(x) = 3/2+sqrt(13)/2 or y-3/2 = -sqrt(13)/2 Raise both sides to the power of two: x = 9 or x = (3/2+sqrt(13)/2)^2 or y-3/2 = -sqrt(13)/2 Add 3/2 to both sides: x = 9 or x = (3/2+sqrt(13)/2)^2 or y = 3/2-sqrt(13)/2 Substitute back for y = sqrt(x): x = 9 or x = (3/2+sqrt(13)/2)^2 or sqrt(x) = 3/2-sqrt(13)/2 Raise both sides to the power of two: x = 9 or x = (3/2+sqrt(13)/2)^2 or x = (3/2-sqrt(13)/2)^2 -3-10 sqrt(x)+x^(3/2) => -3-10 sqrt(9)+9^(3/2) = -6: So this solution is incorrect -3-10 sqrt(x)+x^(3/2) => -3-10 sqrt((3/2-sqrt(13)/2)^2)+((3/2-sqrt(13)/2)^2)^(3/2) = -6: So this solution is incorrect -3-10 sqrt(x)+x^(3/2) => -3-10 sqrt((3/2+sqrt(13)/2)^2)+((3/2+sqrt(13)/2)^2)^(3/2) = 0: So this solution is correct The solution is: Answer: | | x = (3/2+sqrt(13)/2)^2 =~10.908......

x+1/x=?=10.908 + 1/10.908=~11

Feb 8, 2016
#2
+5

when you simplifing that it is clearly shown that x=9

$$x*\sqrt{x}-10*\sqrt{x}-3=0\\(x-10)*\sqrt{x}-3=0\\(x-10)*\sqrt{x}=3\text{ (square both sides)}\\(x-10)^2*x=9\\ x=9$$

$$9+\frac{1}{9}=9.1111111111111111$$

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Feb 8, 2016
#3
+10

Solveit: Substitute your 9 into the equation and see if it balances:

27 - 30 - 3 = -6, Then the answer that I got:

~36 - 33 - 3 =0 !!!!!!!!??????

Feb 8, 2016
#4
0

yeah you are right :( but there s need to be another answer more easier

Feb 9, 2016
#5
0

Thanks guest :)        Yes, this is an interesting question Solveit,

I have put this aside for the next wrap so others can look at it too :)

Feb 14, 2016
#6
+15

Let  sqrt(x) = y, and the first equation becomes

$$\displaystyle y^{3} - 10y - 3 = 0.$$

This clearly has a solution y = -3, so removing the factor (y + 3), the equation can be written

$$\displaystyle (y+3)(y^{2}-3y-1)=0,$$

yielding two further solutions for y,

$$\displaystyle y = (3 \pm\sqrt{13})/2.$$

That suggests three possible values for sqrt(x), but, (and maybe it's implied by the question ?), surely it is reasonable to assume that sqrt(x) is positive ?

If that's the case, then

$$\displaystyle \sqrt{x}=(3+\sqrt{13})/2.$$

Squaring that,

$$\displaystyle x = (11 + 3\sqrt{13})/2$$, so that

$$\displaystyle x + \frac{1}{x}=\frac{11+3\sqrt{13}}{2}+\frac{2}{11+3\sqrt{13}}$$,

$$\displaystyle =\frac{(11+3\sqrt{13})^{2}+2^{2}}{2(11+3\sqrt{13})}=\frac{121+33\sqrt{13}}{11+3\sqrt{13}} = 11.$$

Guest Feb 18, 2016
#7
+5

That is really neat.  Thanks anon.

I wish you had identified yourself.

Is it you , Bertie ://

Feb 19, 2016
#8
+5

Hi Melody

Yes, guilty.

Sorry, l should have signed it.

- Bertie

Feb 19, 2016