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How do you solve # x(x-5)<x(x-4)+2#?

Answer 1

Nathan Axel

#x#>#-2#

#x^2-5x #<# x^2-4x+2#

Subtract #x^2# from both sides

#-5x#<#-4x+2#

Add #5x# to both sides

#0#<#x+2#

Subtract 2 from both sides

#-2#<#x#

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Answer 2

Eli Assouline

To solve the inequality (x(x-5) < x(x-4) + 2), you can start by simplifying both sides of the equation. Then, isolate the variable (x) to determine the range of values that satisfy the inequality.

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Answer 3

Jameson Allen

To solve the inequality (x(x-5) < x(x-4) + 2), we can begin by simplifying both sides:

[x(x-5) < x(x-4) + 2]

Expanding both sides:

[x^2 - 5x < x^2 - 4x + 2]

Now, let's move all terms to one side to set the inequality to zero:

[x^2 - 5x - (x^2 - 4x + 2) < 0]

[x^2 - 5x - x^2 + 4x - 2 < 0]

[ -x + 2 < 0]

Now, let's solve for (x):

[ -x + 2 < 0]

[ -x < -2]

Divide both sides by (-1), remember to reverse the inequality when dividing or multiplying by a negative number:

[x > 2]

So, the solution to the inequality is (x > 2).

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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 # x(x-5)&lt;x(x-4)+2#?

How do you solve # x(x-5)<x(x-4)+2#?