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Minchanka [31]
3 years ago
7

What substitution should be used to rewrite 4x^4-21x^2+20=0 as a quadratic equation

Mathematics
2 answers:
Andreyy893 years ago
8 0

let y = x^2

4y^2 -21y + 20 = 0

solve for y

then substitute back in to get x

azamat3 years ago
7 0

Answer:

4y^2 - 21y + 20 = 0

Step-by-step explanation:

As (4x^4 - 21x^2 + 20=0) being a biquadratic equation, meaning it is a 4-degree equation without the terms of degree 1 and 3.

To solve this particular case you only need to replace the variables of degree 4 and 2 for a new variable of degree 2 and 1 and solve it as regular quadratic equation and then undo the change.

* Using a new variable we can rewrite the equation:

Let y be the new variable;

y= x^2, then y^2= x^4

Then we have;

4y^2- 21y+20=0

Solve for y and then find X which will be the roots for the biquadratic equation.

 

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Step-by-step explanation:

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6 0
3 years ago
Cuales son los terminos que siguen en cada sucesion:a)2,4,8,16,32,64,....B)1,3,5,7,9,....C)1,3,6,10,15,21,28,36,45
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Answer:

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Step-by-step explanation:

a) Para la secuencia 2,4,8,16,32,64, ....

El siguiente número es dos veces el número anterior.

Por ejemplo, 4 = 2 * 2, 8 = 4 * 2, 16 = 8 * 2 etc.

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El siguiente número es igual al número anterior + 2

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c) Para la secuencia1,3, 6,10,15,21,28,36,45

La diferencia entre los números aumenta en +1.

3-1 = 2

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Step-by-step explanation:

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