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fiasKO [112]
3 years ago
7

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

Mathematics
2 answers:
vekshin13 years ago
8 0

Answer is A.

The values of x must be greater than or equal to 1 and the value of x must be less than or equal to -2.

x >= 1.....x <= -2

Please see attachment!


Svetradugi [14.3K]3 years ago
8 0

Answer: The solution set is (-\infty, -2)\cup (1, \infty).

Step-by-step explanation:  We are given to find the solution set of the following quadratic inequality:

x^2+x-2>0.

The solution is as follows:

x^2+x-2>0\\\\\Rightarrow x^2+2x-x-2>0\\\\\Rightarrow x(x+2)-1(x+2)>0\\\\\Rightarrow (x+2)(x-1)>0.

We know that if 'a' and 'b' are two numbers such that a × b > 0, then either both 'a' and 'b' are greater than 0 or both of them are less than 0.

Therefore, we have

either

x+2>0,~~~x-1>0\\\\\Rightarrow x>-2,~~~~x>1\\\\\Rightarrow x>1,

or

x+2

Thus, the required solution set is x < -2 and x > 1. We can also write the solution set in terms of intervals as (-\infty, -2)\cup (1, \infty).

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The dimensions of the rectangle are: b=9.65 cm and h=4.35 cm or b=4.35 cm and h= 9.65 cm.

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The question gives:

Perimeter=28cm

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If the perimeter is the sum of all sides of the rectangle, you have:

Perimeter= 2b+2h=28

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You can write a system of linear equations.

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From equation 2, you have  b=\frac{42}{h}. Then, you can replace it in equation 1.

2*\frac{42}{h}+2h=28 \\ \\ 84+2h^2=28h\\ \\ 2h^2-28h+84=0 \; dividing\; by\; 2\\ \\ h^2-14h+42=0

Now, you should solve the quadratic equation.

h_{1,\:2}=\frac{-\left(-14\right)\pm \sqrt{\left(-14\right)^2-4\cdot \:1\cdot \:42}}{2\cdot \:1}

h_{1,\:2}=\frac{-\left(-14\right)\pm \:2\sqrt{7}}{2\cdot \:1}

h_1=\frac{-\left(-14\right)+2\sqrt{7}}{2\cdot \:1}=7+\sqrt{7}=9.65 cm

h_2=\frac{-\left(-14\right)-2\sqrt{7}}{2\cdot \:1}=7-\sqrt{7}=4.35cm

If h1=9.65 cm, then b_1=\frac{42}{9.65} =4.35 cm, and if  h2=4.35 cm, then b_1=\frac{42}{4.35} =9.65 cm.

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Answer:

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Step-by-step explanation:

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