Question

Не вычисляя корней уравнения x^2-3x-2=0 найдите x1/x2^3 + x2/x1^3

Answer 1

Из уравнения х² - 3х - 2 = 0 по теореме Виета имеем:

{x₁ + x₂ = 3

{x₁ * x₂ = - 2

Найти $\frac{x_1}{x_2^3} +\frac{x_2}{x_1^3}$

1) Упростим

$\frac{x_1}{x_2^3} +\frac{x_2}{x_1^3} =\frac{x_1^4+x_2^4}{x_1^3x_2^3} =\frac{x_1^4+x_2^4}{(x_1x_2)^3}$

2) По теореме Виета

$x_1*x_2 = -2$

Отсюда

$(x_1*x_2)^3=(-2)^3=-8$

3) Осталось найти (х₁⁴ + х₂⁴), для этого воспользуется первым уравнением теоремы Виета

х₁ + х₂ = 3

Возведём обе части в четвёртую степень:

$(x_1+x_2)^4=3^4$

$(x_1^2+2x_1x_2+x_2^2)^2=81$

Так как х₁*х₂ = -2, вместо произведения х₁х₂ подставим (-2) и получим:

$(x_1^2-4+x_2^2)^2=81$

$(x_1^2-4)^2+2(x_1^2-4)*x_2^2+x_2^4=81$

$x_1^4-8x_1^2+16+2x_1^2x_2^2-8x_2^2+x_2^4=81$

$(x_1^4+x_2^4)-8(x_1^2+x_2^2)=81-16-2x_1^2x_2^2$

$x_1^4+x_2^4 - 8(x_1^2+2x_1x_2+x_2^2-2x_1x_2)=65-2(x_1x_2)^2$

$[tex] x_1^4+x_2^4-8*(3^2-2*(-2))=65-2*4$

$x_1^4+x_2^4-8*(9+4)=57$

$x_1^4+x_2^4=57 +104$

$x_1^4+x_2^4=161$

4) Наконец, получим:

$\frac{x_1^4+x_2^4}{(x_1x_2)^3} =\frac{161}{(-2)^3} =-\frac{161}{8} =- 20\frac{1}{8}= -20, 125$

Answer 2

To find the value of the expression x1/x2^3 + x2/x1^3 without calculating the roots of the equation x^2-3x-2=0, we can use Vieta's formulas.

In a quadratic equation of the form ax^2 + bx + c = 0, the sum of the roots is given by -b/a, and the product of the roots is given by c/a.

For the equation x^2 - 3x - 2 = 0, the sum of the roots (x1 + x2) is -(-3)/1 = 3, and the product of the roots (x1 * x2) is (-2)/1 = -2.

Now, let's express x1/x2^3 + x2/x1^3 in terms of the sum and product of the roots:

x1/x2^3 + x2/x1^3 = (x1^4 + x2^4) / (x1^3 * x2^3)

We can then factor the numerator using the fourth power sum identity:

x1^4 + x2^4 = (x1^2 + x2^2)^2 - 2x1^2x2^2

Now, we can express the numerator in terms of the sum and product of the roots:

x1^2 + x2^2 = (x1 + x2)^2 - 2x1x2 = 3^2 - 2*(-2) (Using the values we found earlier) = 9 + 4 = 13

x1^2x2^2 = (x1x2)^2 = (-2)^2 = 4

Therefore,

x1^4 + x2^4 = (x1^2 + x2^2)^2 - 2x1^2x2^2 = 13^2 - 2*4 = 169 - 8 = 161

Now, let's go back to the expression x1/x2^3 + x2/x1^3:

x1/x2^3 + x2/x1^3 = (x1^4 + x2^4) / (x1^3 * x2^3) = 161 / (x1^3 * x2^3)

Now, we can use the product of the roots (x1 * x2) and the sum of the roots (x1 + x2) that we found earlier:

x1^3 * x2^3 = (x1 * x2)^3 = (-2)^3 = -8

Therefore,

x1/x2^3 + x2/x1^3 = 161 / (x1^3 * x2^3) = 161 / (-8) = -20.125

So, the value of x1/x2^3 + x2/x1^3 is -20.125.