(х^2-x+1)(x^2-x-7)=65

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To solve the equation (х^2-x+1)(x^2-x-7) = 65, we can first expand the expression on the left side:

(х^2-x+1)(x^2-x-7) = x^4 - x^3 - 7x^2 + x^3 - x^2 - 7x + x^2 - x + 1 = x^4 - 7x^2 - 7x + 1

Now we set this expression equal to 65:

x^4 - 7x^2 - 7x + 1 = 65

Subtracting 65 from both sides, we get:

x^4 - 7x^2 - 7x + 1 - 65 = 0

x^4 - 7x^2 - 7x - 64 = 0

Now we have a quartic equation that can be difficult to solve. One way to solve this would be to factor the equation or use numerical methods like Newton-Raphson method to find the roots.