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jeka57 [31]
2 years ago
13

What is the solution set of the quadratic inequality x^2+x-2>0

Mathematics
2 answers:
wariber [46]2 years ago
7 0
The photo will explain

STatiana [176]2 years ago
3 0

Answer: (-∞, -2) U (1, ∞)

<u>Step-by-step explanation:</u>

    x² + x - 2 > 0

First, find the zeros by setting the equation EQUAL to zero:

   x² + x - 2 = 0

 (x + 2)(x - 1) = 0

x + 2 = 0    x - 1 = 0

  x = -2        x = 1

Next, choose a test point to the left, between, and to the right of the zeros and check to see if the test points are positive (greater than zero)

Left (x = -3):    (-3 + 2)(-3 - 1) = (-)(-) = +

Between (x = 0):   (0 + 2)(0 - 1) = (+)(-) = -

Right (x = 2):   (2 + 2)(2 - 1) = (+)(+) = +

**************************************************

The left and right test points are positive (greater than zero) so the solution is x < -2 and x > 1

Interval Notation: (-∞, -2) U (1, ∞)



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Answer:

Which equation is true when the value of x is -12    HERE YOU GO!!

Step-by-step explanation:

tricky ... let's see ...

I notice that if we subtract xy from both sides we get

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then

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and

x = 21/7 = 3

so there is only one value of x that satisfies the equation

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Going back to the original equation we see that any value of y will satisfy the original equation

we can see this by rearranging things:

7x + xy - xy = 21

here, I have performed the subtraction of xy on the right side as above, but have left the left side undone

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Note that the above can also be written as

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or

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now, since anything times zero equals zero,

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Let's summarize:

1) x = 3

2) y = anything

looking back at the original question;

1) the equation is true for all ordered pairs

FALSE (only one x works, not all x)

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3) For each value of x, there is one and only one value of y that makes the equation true,

FALSE for each value of x, for the one value of x, x=3, y can be any number, which is an infinite number, not one

4) for each value of y, there is one and only on value of x that makes the equation true.

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A

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