Prove that the equation $x^2 + |x - 1| = 0$ has no real solution.

Question

I was asked to prove that the equation $x^2 + |x - 1| = 0$ has no real solution. I understood that 'no real solution' means no solution in $\Bbb R$.

The problem here is, I have no idea whatsoever how to prove this and I don't know where to start.

An advanced thanks for all the helpful answers. I wish to be able to learn how to prove not only the abovementioned equations, but also other problems that may be given to me.

Answer 1

Hint: You always have $x^2\geqslant 0$. And you always have $|x-1|\geqslant0$ too.

Answer 2

In fact this requires to consider 2 relations within their appropriate realm. $$x^2+x-1=0\ for\ x\ge 1$$ $$x^2-x+1=0\ for\ x<1$$

--- rk

Answer 3

Two cases:

$x-1\ge 0$

Then the equation becomes $x^2+x-1=0$ for which $x = -.5\pm 1.118 \le 1$ and hence no solution because $x \gt 1$

$x-1\le 0$

Then the equation becomes $x^2-x+1=0$ for which x is a complex root

Thus $x^2 +|x-1| = 0$ has no real roots

Answer 4

Observe that for all $x\in\mathbb R,x^2 \geq 0$ and $|x - 1| \geq 0$. Then $x^2 + |x - 1| = 0 \iff x^2=0$ and $|x - 1| = 0\iff x=0$ and $x=1\iff x\in\varnothing$.

Answer 5

More general: if $a,b \ge 0$ and $a+b=0$, then we have $a=b=0$.

In your case we get $x^2=|x-1|=0$, but this is impossible !

Answer 6

$|x^2| +|x-1| = 0$.

$\iff$

$ |x^2| =0$ and(!) $|x-1|=0$.

Hence?

Answer 7

When $x=0$, then $x^2 + |x - 1|=1$

When $x=1$, then $x^2 + |x - 1| = 1$

Otherwise, $x^2 > 0$ and $|x-1| > 0$. Thus $x^2 + |x - 1| > 0$.