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Darya [45]
3 years ago
7

Find the value or values of x in the quadratic equation x2=6x-9.

Mathematics
2 answers:
nordsb [41]3 years ago
8 0

Hi Harris


x²=6x-9

First thing we need to do is

subtract 6x-9 from the booth sides

x²-(6x-9)=6x-9-(6x-9)

x²-6x+9=0

Now we gonna factor the left side

(x-3)(x-3)=0

Set factor equal 0

x-3=0 or x-3=0

x=3


I hope that's help !

ozzi3 years ago
5 0

x² = 6x - 9, or x² - 6x + 9 = 0, or x² - 2*x*3 + 3² = 0, or (x-3)² = 0 => x₁ = x₂ = 3.

Green eyes.

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There is a multiple zero at 0 (which means that it touches there), and there are single zeros at -2 and 2 (which means that they cross). There is also 2 imaginary zeros at i and -i.


You can find this by factoring. Start by pulling out the greatest common factor, which in this case is -x^2.


-x^6 + 3x^4 + 4x^2

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Now we can factor the inside of the parenthesis. You do this by finding factors of the last number that add up to the middle number.


-x^2(x^4 - 3x^2 - 4)

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Now we can use the factors of two perfect squares rule to factor the middle parenthesis.


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We would also want to split the term in the front.


-x^2(x - 2)(x + 2)(x^2 + 1)

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Now we would set each portion equal to 0 and solve.


First root

x = 0 ---> no work needed


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Fifth and Sixth roots

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