What Went Wrong with Solving a Simple Inequality?

[Saitama]

Homework Statement

$$3\sqrt{x}-\sqrt{x+3}>1$$

Homework Equations

The Attempt at a Solution

As obvious from the given inequality, x must be greater than zero.
Rearranging and squaring both the sides,
$$9x>1+x+3+2\sqrt{x+3} \Rightarrow 4x-2>\sqrt{x+3}$$
Squaring again,
$$16x^2+4-16x>x+3 \Rightarrow 16x^2-17x+1>0 \Rightarrow (x-1/16)(x-1)>0$$
Hence, $x \in (0,1/16) \cup (1,\infty)$ but this is wrong. The correct answer is $(1,\infty)$. I don't see what I did wrong.

Any help is appreciated. Thanks!

 

[Curious3141]

Homework Statement

$$3\sqrt{x}-\sqrt{x+3}>1$$

Homework Equations

The Attempt at a Solution

As obvious from the given inequality, x must be greater than zero.
Rearranging and squaring both the sides,
$$9x>1+x+3+2\sqrt{x+3} \Rightarrow 4x-2>\sqrt{x+3}$$
Squaring again,
$$16x^2+4-16x>x+3 \Rightarrow 16x^2-17x+1>0 \Rightarrow (x-1/16)(x-1)>0$$
Hence, $x \in (0,1/16) \cup (1,\infty)$ but this is wrong. The correct answer is $(1,\infty)$. I don't see what I did wrong.

Any help is appreciated. Thanks!

This is a "classic" elementary mistake in algebra. When you square (or more generally, take even powers) of equations, you often introduce "redundant roots". These roots solve the final form you get, but are not solutions to the original equation. The basic reason is that the square of both $a$ and $-a$ give you the same value $a^2$.

Try putting $x = \frac{1}{16}$ in the original LHS and see what the expression evaluates to. Do you see what's going on?

In the future, when you square equations to solve them (if it's unavoidable, as in this case), always go back and test your solution set using the original equation. Discard solutions that don't satisfy the original equation.

 

[Saitama]

This is a "classic" elementary mistake in algebra. When you square (or more generally, take even powers) of equations, you often introduce "redundant roots". These roots solve the final form you get, but are not solutions to the original equation.

Try the value 1/16 in the original LHS and see what RHS becomes. Do you see what's going on?

I understand what you say but is it possible to be able to point out the redundant interval immediately? I mean, in this case, I had to substitute 1/16 and found that it did not satisfy the inequality. How should I go about selecting the "redundant" roots?

 

[Curious3141]

I understand what you say but is it possible to be able to point out the redundant interval immediately? I mean, in this case, I had to substitute 1/16 and found that it did not satisfy the inequality. How should I go about selecting the "redundant" roots?

Generally, you test the solutions at the end and decide which to accept. I'm not familiar with any simpler method.

 

[Saitama]

Generally, you test the solutions at the end and decide which to accept. I'm not familiar with any simpler method.

Thanks!

 

[Curious3141]

Thanks!

Sure, and thanks for the thanks!

 

[ehild]

Homework Statement

$$3\sqrt{x}-\sqrt{x+3}>1$$

Homework Equations

The Attempt at a Solution

As obvious from the given inequality, x must be greater than zero.
Rearranging and squaring both the sides,
$$9x>1+x+3+2\sqrt{x+3} \Rightarrow 4x-2>\sqrt{x+3}$$

Before "squaring again", you have to investigate the equation. If 4x-2>0, you can proceed. But 4x-2<0 is not possible, as the square root is defined as non-negative number. A negative number can not be greater than a non-negative one. Therefore x must be greater then 0.5.


ehild

 

[Saitama]

Before "squaring again", you have to investigate the equation. If 4x-2>0, you can proceed. But 4x-2<0 is not possible, as the square root is defined as non-negative number. A negative number can not be greater than a non-negative one. Therefore x must be greater then 0.5.


ehild

Nice catch ehild, thank you!