Mirror mirror on the wall
Algebra
Level
3
Consider the equation above. Which of the following pairs show the correct values of ?
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1 solution
Let's consider p(x)= 2x^4 - 11x^3 + 16x^2 - 11x + 2 = 0. Then x=0 isn't a root of p(x). This implies that we can divide for x^2;
2x^2 - 11x + 16 - 11/x + 2/x^2 = 0;
2(x^2 + 1/x^2) - 11(x + 1/x) + 16 = 0;
2((x + 1/x)^2 - 2) - 11(x + 1/x) + 16 = 0; Now, we call y = x + 1/x. Therefore,
2(y^2 -2) - 11y + 16 = 0;
2y^2 - 11y + 14 = 0;
y = x + 1/x = 3/2 or 8