Question

 Using Intermediate Value Theorem, show that

x2 = √𝑥 + 1 has a root in (1,2)

Answer 1

Solution

Given that

$x^2=\sqrt{x+1}$

This can be re-written as

$f(x)=x^2-\sqrt{x+1}$ where $f(x)=0$ will give us the root(s) of the equation.

Now for

$x=1$ $f(1)=(1)^2-\sqrt{(1)+1}=-0.4142... <0$

And

$x=2$ $f(2)=(2)^2-\sqrt{(2)+1}=2.2679... >0$

Since $f(1)<0$ and $f(2)>0$

Therefore, $f(x)=0$ for some value of $x$ which lies with in the interval $(1, 2)$

Hence $x^2=\sqrt{x+1}$ has a root within the interval $(1, 2)$

The plot below shows that $x^2=\sqrt{x+1}$ has root between $(1, 2)$ , which is $x=1.221$