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Elis [28]
3 years ago
11

What are the real and complex solutions of the polynomial equation? x^4-41x^2=-400

Mathematics
1 answer:
tankabanditka [31]3 years ago
7 0
<h3>Answer:</h3>

The real solutions are -5, -4, 4, 5. There are no complex solutions.

<h3>Step-by-step explanation:</h3>

The equation ...

... x^4 -41x^2 +400 = 0

can be factored as ...

... (x^2 -16)(x^2 -25) = 0

... (x -4)(x +4)(x -5)(x +5) = 0

So, all roots are real and are ...

... x ∈ {-5, -4, 4, 5}

_____

These are the values of x that make the factors zero.

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4 - 7x = 1 - 6x <br>what is that​
DanielleElmas [232]

Answer:

x = 3

Step-by-step explanation:

4 - 7x = 1 - 6x (Given)

4 - 1 - 7x = 1 - 1 - 6x (Subtraction Property of Equality)

3 - 7x = -6x (Simplify)

3 - 7x + 7x = -6x + 7x (Addition Property of Equality)

3 = x (Simplify)

x = 3 (Symmetric Property of Equality)

4 0
3 years ago
What is the answer to (9w4r6)2
oksian1 [2.3K]

Answer:

the answer to (9w4r6)2 is 18w^{4}r^{6}

6 0
3 years ago
Scive fer x<br>COR<br>how can i possibly solve this??
frez [133]
The equation is
(5x + 2) + (16x + 10) = 180
8 0
3 years ago
Plz help me!!!!!!!!!!!!
Mashutka [201]

Answer:  3, -3, 3i, -3i

<u>Step-by-step explanation:</u>

<u>x^4-81=0\\\\Factor:\\(x^2-9)(x^2+9)=0\\(x-3)(x+3)(x^2+9)=0\\\\\text{Apply the Zero Product Property:}\\x-3=0\qquad x+3=0\qquad x^2+9=0\\\boxed{x=3}\qquad \qquad \boxed{x=-3}\qquad \quad x^2=-9\\.\qquad \qquad \qquad \qquad \qquad \qquad x=\sqrt{-9}\\.\qquad \qquad \qquad \qquad \qquad \qquad \boxed{x=\pm 3i}</u>

7 0
3 years ago
Help math math math
lutik1710 [3]

Hello.

In order to convert this equation into slope intercept form, we need to divide the entire equation by 7:

\tt{8x-7y=13

\tt{\displaystyle\frac{8x}{7} -\frac{7y}{7}=\frac{13}{7}

Simplify:

\tt{\displaystyle\frac{8x}{7}-y=\frac{13}{7}

Add y to both sides:

\Large\boxed{\sf{y=\displaystyle\frac{8x}{7} -\frac{13}{7}}}

I hope it helps.

Have an outstanding day. :)

\boxed{imperturbability}

5 0
2 years ago
Read 2 more answers
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