Simple equation help

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July 30th 2008, 06:32 AM
Jones

Simple equation help

Hi, im having problems with a simple equation that looks like this:
$\sqrt{x+3} = \sqrt{x-2} + \sqrt{x-5}$

After squaring both sides i have: $x+3=2x-7 +2\sqrt{x-2}\cdot \sqrt{x-5} \rightarrow x-10+2 \cdot (x-2)^\frac{1}{2} \cdot (x-5) ^\frac {1}{2}$ Wich gives: $x-10+2x^{2}-14x+20=0$

However, the anser should be 6 and i get 4. Im not sure about what i am doing wrong =/
July 30th 2008, 07:11 AM
Serena's Girl

When you multiplied sqrt(x - 2) and sqrt(x - 5) to each other, you forgot to include the square root sign in the product.

You ended up with a final expression that doesn't include any square root signs so it became inaccurate.
July 30th 2008, 07:22 AM
Chris L T521

Quote:

Originally Posted by Jones

Hi, im having problems with a simple equation that looks like this:
$\sqrt{x+3} = \sqrt{x-2} + \sqrt{x-5}$

After squaring both sides i have: $x+3=2x-7 +2\sqrt{x-2}\cdot \sqrt{x-5} \rightarrow x-10+2 \cdot (x-2)^\frac{1}{2} \cdot (x-5) ^\frac {1}{2}$ Wich gives: $x-10+2x^{2}-14x+20=0$

However, the anser should be 6 and i get 4. Im not sure about what i am doing wrong =/

Starting from this step:

$x-10+2 \cdot (x-2)^\frac{1}{2} \cdot (x-5) ^\frac {1}{2}=0$

This becomes: $\sqrt{x^2-7x+10}=\tfrac{1}{2}(10-x)$

Now square both sides:

$x^2-7x+10=\tfrac{1}{4}(x^2-20x+100)$

$\implies 4x^2-28x+40=x^2-20x+100$

$\implies 3x^2-8x-60=0$

$\implies (x-6)(3x+10)=0$

$\implies \color{red}\boxed{x=6} \ \text{or} \ \color{red}\boxed{x=-\tfrac{10}{3}}$

Test values:

$\sqrt{-\tfrac{10}{3}+3} = \sqrt{-\tfrac{10}{3}-2} + \sqrt{-\tfrac{10}{3}-5}$

$\sqrt{\tfrac{1}{3}}i=4\sqrt{\tfrac{1}{3}}i+5\sqrt{ \tfrac{1}{3}}i$

This doesn't work...

$\sqrt{6+3} = \sqrt{6-2} + \sqrt{6-5}$

$\implies 3=2+1 \implies 3=3$

$x=6$ is the only solution.
July 30th 2008, 07:23 AM
Jones

Quote:

Originally Posted by Serena's Girl

When you multiplied sqrt(x - 2) and sqrt(x - 5) to each other, you forgot to include the square root sign in the product.

You ended up with a final expression that doesn't include any square root signs so it became inaccurate.

Since we have $\sqrt(x-2) \cdot \sqrt(x-5)$
Its the same thing as $(x-2)^\frac{1}{2} \cdot (x-5)^\frac{1}{2}$

Accordign to the exponetial rules $A^b \cdot B^a = AB^{a + b}$ and $\frac{1}{2} + \frac {1}{2} = 1$

Therefore $((x-2) \cdot (x-5))^{1}$
July 30th 2008, 07:26 AM
Chris L T521

Quote:

Originally Posted by Jones

Since we have $\sqrt(x-2) \cdot \sqrt(x-5)$
Its the same thing as $(x-2)^\frac{1}{2} \cdot (x-5)^\frac{1}{2}$

Accordign to the exponetial rules $A^b \cdot B^a = AB^{a + b}$ and $\frac{1}{2} + \frac {1}{2} = 1$

Therefore $((x-2) \cdot (x-5))^{1}$

But you can only do that if the exponential functions have the same base.

In the case of square roots, $\sqrt{a}\cdot\sqrt{b}=\sqrt{ab} \ \forall a\geq 0, \ b\geq 0$

--Chris
July 30th 2008, 07:28 AM
Jones

Oh, right.

Thank you (Bow)