How do you solve and graph #x^2+1

Question

How do you solve and graph #x^2+1<2x#?

Sophie Adams · Accepted Answer

#x>1#

We start with:

#x^2+1<2x#

We'll solve this the same way as if it were an equal sign - so let's drop in an equal sign for now, so:

#x^2-2x+1=0#

#(x-1)(x-1)=0# only need to find the one solution, so

#x-1=0#

#x=1#

Ok - so we know the x value where the two sides are equal. So where are the values of x that satisfy the inequality - are they to the left or to the right of 1? Let's test what happens when #x=0#:

#x^2+1<2x#

#0+1<2(0)#

#1<0# - No. So it's not values less than 1 that will work - it's values more than one. So the solution is:

#x>1#

For the graph, it'll be a ray along the x-axis (or number line) starting at #x=1# with a little circle around that point (to indicate that #x=1# is not a solution) and the ray pointing to the right (towards larger numbers).

Sadie Alexander · Answer

undefined To solve the inequality $x^2 + 1 < 2x$:

1. Subtract $2x$ from both sides to get $x^2 - 2x + 1 < 0$.
2. Factor the quadratic expression to get $(x - 1)^2 < 0$.
3. Recognize that a square is always non-negative, so the expression $(x - 1)^2$ is always greater than or equal to zero.
4. Since we're looking for where $(x - 1)^2$ is less than zero, there are no real solutions.

To graph the inequality:

1. Plot the parabola $y = x^2 - 2x + 1$.
2. Since the expression $(x - 1)^2$ is always greater than or equal to zero, the region where $(x - 1)^2 < 0$ does not exist on the real number line.
3. Therefore, the graph of the inequality $x^2 + 1 < 2x$ is an empty set on the real number line.

How do you solve and graph #x^2+1&lt;2x#?

How do you solve and graph #x^2+1<2x#?