Problems in $\sqrt{5x+4}=x-2$

Question

So when I solve $\sqrt{5x+4}=x-2$, I end up with $x(x-9)=0$. Yet only when $x=9$ is the original inequality satisfied. Can somebody give me some details on what exactly goes wrong here?

$a^2=b^2$ does not imply $a=b$. – Angina Seng Aug 26 '19 at 18:22 — Angina Seng, Aug 26 '19 at 18:22
Including how you solved it would help. – The Count Aug 27 '19 at 00:12 — The Count, Aug 27 '19 at 00:12

score 4 · Accepted Answer · answered Aug 26 '19 at 18:19

I guess you started by squaring both sides:

$$\sqrt{5x+4} = x+2 \tag{A}$$

$$5x+4 = (x+2)^2 \tag{B}$$

While it is true that $(A)$ implies $(B)$, the other direction does not hold. Maybe a simple example is following: While $3^2 = 9 = (-3)^2$ is true, $3 = -3$ is clearly not true.

So if some $x = x_0 $ satisfis equation $(A)$ it must also satisfy the second equation $(B)$, but if some $x = x_1$ satisfies equation $(B)$ it does not necessarily satisfy $(A)$.

score 2 · Answer 2 · answered Aug 26 '19 at 18:20

If you insert $x=0$, you can see what goes on: $$ \sqrt{5\cdot0+4}=2\\ 0-2=-2 $$ These are different. However, the moment you square both sides, you don't see this difference any more, as both sides become $4$.

In general, when you square an equation, in addition to the actual solutions where the left-hand side is equal to the right-hand side of the original equation, you suddenly include the cases where the left-hand side is the negative of the right-hand side.

score 1 · Answer 3 · answered Aug 26 '19 at 18:20

It is just because$$\sqrt{5x+4}=x-2\implies5x+4=(x-2)^2,\tag1$$but this is not an equivalence. However, it is clear that, if $x<2$, then the equality $\sqrt{5x+4}=x-2$ cannot hold (since $\sqrt{5x+4}\geqslant0$ and $x-2<0$). And if you restrict $x$ to $[2,\infty)$ then, yes, $(1)$ is an equivalence.

score 0 · Answer 4 · answered Aug 26 '19 at 18:20

0

At $x=0$, the LHS is $2$, while the RHS is $-2$. So the new solution come from squaring both sides, because $(2)^2=(-2)^2$.

answered Aug 26 '19 at 18:20

Botond

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score 0 · Answer 5 · answered Aug 26 '19 at 18:21

When you square both sides to solve, you create an extra incorrect answer. See, at x=0, the squares of both sides of the equation are equal, but the sides of the original equation are not. The constraint that x-2 cannot be negative is lost in the process of squaring, and thus an extra solution is created.

score 0 · Answer 6 · answered Aug 26 '19 at 18:43

As a general rule, when you are trying to solve an equation with radicals in the form $\sqrt{P(x)}>Q(x)$, you have to solve the following system: $$\left\{\begin{matrix} P(x)\geq0 \\Q(x)\geq0 \\P(x)=Q(x)^2 \end{matrix}\right.$$

So, I have:
$$\left\{\begin{matrix} 5x+4\geq0 \\x-2\geq0 \\5x+4=x^2-4x+4 \end{matrix}\right.$$

The solutions of the system are: $x(x-9)=0$ but $x=0$ isn't acceptable, so the only correct is $x=9$.

Problems in $\sqrt{5x+4}=x-2$

6 Answers6