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miss Akunina [59]
3 years ago
12

Find the difference 7/x^2-3x - 1/x^2-9

Mathematics
1 answer:
Goshia [24]3 years ago
6 0
To solve this problem you must apply the proccedure shown below:

 1. You have the following expression given in the problem above:

 <span>7/x^2-3x - 1/x^2-9
</span>
 2. When you factor the denominator, you obtain:

 <span>7/x(x-3) - 1/(x+3)(x-3)
 

 3. By simplifying the expression, you obtain:

 3(2x+7)/x(x+3)(x-3)
</span> 3(2x+7)/x(x^2-9)

 The answer is: 
 3(2x+7)/x(x^2-9)
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Answer:

The fundamental theorem of algebra tells you that the equation will have two complex roots since the degree of the polynomial is 2. The roots are x=1\pm i\sqrt{7}.

Step-by-step explanation:

Consider the provided information.

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2x^2-4x+16=0

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Now find the root of the equation.

For the quadratic equation of the form ax^2+bx+c=0 the solutions are: x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

Substitute a=2,\:b=-4,\:\ and \ c=16 in above formula.

x_{1,\:2}=\frac{-\left(-4\right)\pm \sqrt{\left(-4\right)^2-4\cdot \:2\cdot \:16}}{2\cdot \:2}

x_{1,\:2}=\frac{4\pm \sqrt{16-128}}{4}

x_{1,\:2}=\frac{4\pm \sqrt{-112}}{4}

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Hence, the fundamental theorem of algebra tells you that the equation will have two complex roots since the degree of the polynomial is 2. The roots are x=1\pm i\sqrt{7}.

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