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3 years ago

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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Answer:

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Step-by-step explanation:

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3 years ago

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kozerog [31]

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2 years ago

HELP PLEASEE!!! I just need help with the 2 boxes

Soloha48 [4]

Answer:

$y= x - 300$

$y = -246$ -- Her equivalent weight on earth

Step-by-step explanation:

Given

The attached table

Required

Fill in the gaps

(a) The equation

First, we calculate the slope (m)

$m = \frac{y_2 - y_1}{x_2 - x_1}$

From the table, we can say that:

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So, we have:

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$m = \frac{0 +300}{300}$

$m = \frac{300}{300}$

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The equation is then calculated using:

$y - y_1 = m(x - x_1)$

This gives:

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$y +300 = x$

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(b) When x = 54

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