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Solve the system of quadratic inequalities by pure algebraic method
1) x^2 - 7x+10 ≥ 0
2) x^2 +4x-12<0

1) x ≤ 2 or x ≥ 5

2) -6 < x < 2

Step-by-step explanation:

1) We have x^2 - 7x + 10, so let's factor this as if this were a regular equation:

x^2 - 7x + 10 = (x - 2)(x - 5)

So, we now have (x - 2)(x - 5) ≥ 0

Let's imagine this as a graph (see attachment). Notice that the only place that is above the number line is considered greater than 0, and that's when x ≤ 2 or x ≥ 5 (the shaded region).

2) Again, we have x^2 + 4x - 12, so factor this as if this were a regular equation:

x^2 + 4x - 12 = (x + 6)(x - 2)

So now we have (x + 6)(x - 2) < 0

Now imagine this as a graph again (see second attachment). Notice that the only place that is below 0 (< 0) is when -6 < x < 2 (the shaded region).

Hope this helps!

1. x ≤ 2 or x ≥ 5

2. -6 < x < 2

Step-by-step explanation:

1) x^2 - 7x+10 ≥ 0

Find the roots

x² - 7x + 10 = 0

x² - 5x - 2x + 10 = 0

x(x - 5) - 2(x - 5) = 0

(x - 2)(x - 5) = 0

x = 2, 5

It's an "or" case.

x ≤ 2 or x ≥ 5

2) x^2 +4x-12<0

Find the roots

x² + 4x - 12 = 0

x² + 6x - 2x - 12 = 0

x(x + 6) - 2(x + 6) = 0

(x - 2)(x + 6) = 0

x = 2, -6

It's an "and" case.

-6 < x < 2

answer rather than having to add and subtract the measurements of the semi-circle, notice that the semicircle to the left of ad is actually identical in area to the blank semicircle-shaped space to the left of bc. so if we transferred the red semicircle into the blank space, we would end up with a square of side length 24 cm. the area of a square is s^2 = 24^2 = 676 cm^2.

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