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BaLLatris [955]
3 years ago
6

Help please!!!

Mathematics
1 answer:
katovenus [111]3 years ago
4 0

Answer:

x = \frac{7+\sqrt{47}\times i }{4}

Step-by-step explanation:

<u>To solve quadratic systems,we always substitute one variable in terms if the other and then solve the equation.</u>

x + 2y = 6                                 ---------------(1)

y - 5 = (x-2)^{2}         ---------------(2)

y = (x-2)^{2} + 5         ---------------(3)

Substitute (3) in (1) ,

x + 2( (x-2)^{2} + 5 ) = 6

(a + b)^{2} =a^{2} + 2ab + b^{2}

x + 2( x^{2} - 4x + 4 + 5 ) = 6

2x^{2} - 7x + 12=0      --------------(4)

The roots of the quadratic equation ax^{2}  +bx+c is

x = \frac{(-b) + \sqrt{(-b)^{2}-4 \times ac }  }{2 \times a}  -----------(5)

According to equation (5),solution of (4) is

x =  \frac{7 + \sqrt{(-7)^{2}-4 \times 24 }  }{2 \times 2}

x =  \frac{7+\sqrt{49-96}}{4}

x = \frac{7+\sqrt{47}\times i }{4}

 

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From the graph, we have the following observations:

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This means that the graph has a multiplicity of 2 at x = 2 and a multiplicity of 3 at x = -5

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