3 years ago

Help me, please I could fail

Mathematics

1 answer:

balu736 [363]3 years ago

5 0

Answer: I believe the answer is B 8/25.12

Step-by-step explanation:

Tell me if this helps

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The answer is 0=0, 2=5, 4=10, 6= 15. Sorry if I am late. But I hope this helps, and Have an amazing day!!

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P(x) = x3 - 2x2 + 2x<br> What are the real and complex zeros of the equation

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Given:

The polynomial is

$p(x)=x^3-2x^2+2x$

To find:

The real and complex zeros of the equation.

Solution:

We have,

$p(x)=x^3-2x^2+2x$

For zeros, p(x)=0.

$x^3-2x^2+2x=0$

$x(x^2-2x+2)=0$

$x=0\text{ and }x^2-2x+2=0$

The real value of x is 0. The equation $x^2-2x+2=0$ will give complex roots. Here, a=1, b=-2 and c=2.

Using quadratic formula, we get

$x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$

$x=\dfrac{-(-2)\pm \sqrt{(-2)^2-4(1)(2)}}{2(1)}$

$x=\dfrac{2\pm \sqrt{4-8}}{2}$

$x=\dfrac{2\pm \sqrt{-4}}{2}$

On further simplification, we get

$x=\dfrac{2\pm \sqrt{-1}\sqrt{4}}{2}$

$x=\dfrac{2\pm 2i}{2}$

$x=\dfrac{2(1\pm i)}{2}$

$x=1\pm i$

Therefore, the real zero is 0 and the complex zeros are 1+i and 1-i.

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3 years ago