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3 years ago

Solve the equation... 10x - 12 = 48 - 2x

1 answer:

hjlf3 years ago

4 0

If you would like to solve the equation 10 * x - 12 = 48 - 2 * x, you can calculate this using the following steps:

10 * x - 12 = 48 - 2 * x
10 * x + 2 * x = 48 + 12
12 * x = 60 /12
x = 60 / 12
x = 5

The correct result is 5.

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2 years ago

Find all the zeros of the polynomial function p(x) = x3 – 5x2 + 33x – 29

hram777 [196]

Answer:

$\large \boxed{\sf \ \ x=1, \ \ x=2+5i, \ \ x=2-5i \ \ }$

Step-by-step explanation:

Hello,

I assume that we are working in $\mathbb{C}$ , otherwise there is only one zero which is 1. Please consider the following.

First of all, <u>we can notice that 1 is a trivial solution</u> as

$p(1) = 1^3-5\cdot 1^2 + 33\cdot 1-29=1-5+33-29=0$

It means that (x-1) is a factor of p(x) so we can find two real numbers, a and b, so that we can write the following.

$p(x)=(x-1)(x^2+ax+b)=x^3+ax^2+bx-x^2-ax-b=x^3+(a-1)x^2+(b-a)x-b$

Let's identify like terms as below.

a-1 = -5 <=> a = -5 + 1 = -4

b-a = 33

-b = -29 <=> b = 29

$\boxed{ \ p(x)=(x-1)(x^2-4x+29) \ }$

Now, we need to find the zeroes of the second factor, meaning finding x so that:

$x^2-4x+29=0 \ \text{ complete the square, 29 = 25 + 4} \\ \\ x^2-2\cdot 2 \cdot x+2^2+25=0 \\ \\ (x-2)^2=-25=(5i)^2 \ \text{ take the root } \\ \\x-2=\pm 5i \ \text{ add 2 } \\ \\ x = 2+5i \ \text{ or } \ x = 2-5i$

Hope this helps.

Do not hesitate if you need further explanation.

Thank you

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3 years ago