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How do you solve and write the following in interval notation: #x^2 + 6 <2x#?

Answer 1

Julia Adams

The solution #S={O/}#

Let's rewrite the equation

#x^2-2x+6<0#

We complete the squares

#x^2-2x+1+5<0#

#(x-1)^2+5<0#

Let #f(x)=(x-1)^2+5#

So,

#AA x in RR, f(x)>0#

There is no solution to the inequality

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Answer 2

Julia Adams

To solve (x^2 + 6 < 2x):

Move all terms to one side of the inequality: (x^2 - 2x + 6 < 0).
Factor the quadratic equation: ((x - 3)(x - 2) < 0).
Determine the critical points by setting each factor equal to zero: (x - 3 = 0) and (x - 2 = 0).
Solve for x: (x = 3) and (x = 2).
Use a number line to test intervals created by the critical points.
Test a value less than 2, between 2 and 3, and greater than 3 into the original inequality.
Determine which intervals satisfy the inequality.
Write the solution in interval notation.

The solution in interval notation is ((2, 3)).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

How do you solve and write the following in interval notation: #x^2 + 6 &lt;2x#?

How do you solve and write the following in interval notation: #x^2 + 6 <2x#?