Why this value is not the second root of the equation?

Question

$\sqrt{x+3}=x-2$

Why $\frac{5}{2} - \frac{\sqrt{21}}{2}$ is not root?

There is only one restriction: $\sqrt{x+3}$, but $\frac{5}{2} - \frac{\sqrt{21}}{2} > 0$.

$x^2-4x+4=x+3$, $x^2-5x+1=0$, $D=25-4=21>0$

$D>0$, =>, $x = \frac{ 5 \pm \sqrt{21} }{ 2 }$

"There is only one restriction" No, there are two. 1) $\sqrt{x+3}$ is meaningful $\implies x \ge -3$. And $\sqrt{x+3} \ge 0 \implies x+2 = \sqrt{x+3} \ge 0$. — fleablood, May 12 '18 at 22:24

score 8 · Answer 1 · answered May 12 '18 at 22:21

8

There is actually a second restriction to your equation, which is "hidden":

Since $\sqrt{x+3} \geq 0$ you must also have $x-2 \geq 0$, which your root fails.

Note that the second root actually satisfies $\sqrt{x+3}=-(x-2)$.

answered May 12 '18 at 22:21

N. S.

132,525

score 7 · Accepted Answer · answered May 12 '18 at 22:20

7

Because $\left(\frac{5}{2} - \frac{\sqrt{21}}{2}\right)-2$ is negative, so it cannot be the square root of something.

answered May 12 '18 at 22:20

Arnaud Mortier

27,276

score 3 · Answer 3 · answered May 12 '18 at 22:23

3

That is because the irrational equation: $$\sqrt A=B\iff A=B^2\;\textbf{ and }\, B\ge 0,$$ i.e., in the present case, $\;x\ge 2$, whereas $\;\dfrac{5-\sqrt{21}}2<\dfrac12$.

answered May 12 '18 at 22:23

Bernard

175,478

Why this value is not the second root of the equation?

3 Answers3