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# Math Help

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Given that x is an integer such thatx * √x - 5x - 9 * √x = 35 , find x.

Jul 2, 2018

### 2+0 Answers

#1
+1

The solution to this is quite involved:
Solve for x:
-9 sqrt(x) - 5 x + x^(3/2) = 35

Subtract 35 from both sides:
-35 - 9 sqrt(x) - 5 x + x^(3/2) = 0

Simplify and substitute y = sqrt(x).
-35 - 9 sqrt(x) - 5 x + x^(3/2) = -35 - 9 sqrt(x) - 5 (sqrt(x))^2 + (sqrt(x))^3
= y^3 - 5 y^2 - 9 y - 35:
y^3 - 5 y^2 - 9 y - 35 = 0

The left hand side factors into a product with two terms:
(y - 7) (y^2 + 2 y + 5) = 0

Split into two equations:
y - 7 = 0 or y^2 + 2 y + 5 = 0

Add 7 to both sides:
y = 7 or y^2 + 2 y + 5 = 0

Substitute back for y = sqrt(x):
sqrt(x) = 7 or y^2 + 2 y + 5 = 0
Raise both sides to the power of two:
x = 49 or y^2 + 2 y + 5 = 0

Subtract 5 from both sides:
x = 49 or y^2 + 2 y = -5

Add 1 to both sides:
x = 49 or y^2 + 2 y + 1 = -4

Write the left hand side as a square:
x = 49 or (y + 1)^2 = -4

Take the square root of both sides:
x = 49 or y + 1 = 2 i or y + 1 = -2 i

Subtract 1 from both sides:
x = 49 or y = -1 + 2 i or y + 1 = -2 i
Substitute back for y = sqrt(x):
x = 49 or sqrt(x) = -1 + 2 i or y + 1 = -2 i

Raise both sides to the power of two:
x = 49 or x = -3 - 4 i or y + 1 = -2 i

Subtract 1 from both sides:
x = 49 or x = -3 - 4 i or y = -1 - 2 i

Substitute back for y = sqrt(x):
x = 49 or x = -3 - 4 i or sqrt(x) = -1 - 2 i

Raise both sides to the power of two:
x = 49 or x = -3 - 4 i or x = -3 + 4 i

-9 sqrt(x) - 5 x + x^(3/2) ⇒ -9 sqrt(-4 i - 3) - 5 (-3 - 4 i) + (-3 - 4 i)^(3/2) = -5 + 40 i:
So this solution is incorrect
-9 sqrt(x) - 5 x + x^(3/2) ⇒ -9 sqrt(4 i - 3) - 5 (-3 + 4 i) + (-3 + 4 i)^(3/2) = -5 - 40 i:
So this solution is incorrect

-9 sqrt(x) - 5 x + x^(3/2) ⇒ -9 sqrt(49) - 5 49 + 49^(3/2) = 35:
So this solution is correct

The solution is:
x = 49

Jul 2, 2018
#2
+103676
+1

Given that x is an integer such thatx * √x - 5x - 9 * √x = 35 , find x.

$$x\sqrt x - 5x - 9\sqrt x = 35\qquad x\ge 0\\ let\;\;y=\sqrt x\qquad \qquad \qquad y\ge 0\\ y^2*y-5y^2-9y=35 \\ y^3-5y^2-9y=35 \\ y^3-5y^2-9y-35 =0\\$$

I am told that there is an integer solution to this so how do I find it ?

Well I know y must be positive, that makes it easier.

The solution must be a factor of 35.

So it has to be 1 or 5 or 7 or 35.

1-5-9-35 = -48    so 1 is not right

5^3-5*5^2-9*5-35 = -80 so 5 is not correct

7^3-5*7^2-9*7-35 = 0    so 7 is a root

35^3-5*35^2-9*35-35 = 36400 so 35 is not a root.

So the only poisitive integer solution is y=7

so

$$x=y^2 \\x= 49$$

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Graphical check :

When I graphed  $$y^3-5y^2-9y-35=0$$

this is the the result ... just remember that the axis have been reversed.

I know this is a cubic so the cruve cannot turn around again so the only answer is y=7

Which means that x=49.

Jul 2, 2018