Question

If A and B are the root of the equation x2 - 6x + 4 =
0. Find the value of A/B+B/A+2(1/A+1/B)+3AB​

Answer 1

Answer:

The value of the expression is 22.

Step-by-step explanation:

The equation provided is: $x^{2}-6x+4=0$

This is a quadratic equation. The roots of a quadratic equation can be computed using the formula:

$x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$

The values of a, b and c are:

a = 1, b = -6 and c = 4.

The roots are:

$x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}=\frac{-(-6)\pm\sqrt{(-6)^{2}-(4\times1\times4)}}{2\times1} =\frac{6\pm2\sqrt{5}}{2}=3\pm\sqrt{5}$

The value of A and B are $(3-\sqrt{5})$ and $(3+\sqrt{5})$ receptively.

Compute the value of the expression given as follows:

$\frac{A}{B}+\frac{B}{A}+2[\frac{1}{A}+\frac{1}{B}]+3AB=\frac{3-\sqrt{5}}{3+\sqrt{5}} +\frac{3+\sqrt{5}}{3-\sqrt{5}}+2[\frac{1}{3-\sqrt{5}}+\frac{1}{3+\sqrt{5}}]+3(3-\sqrt{5})(3+\sqrt{5})\\=[\frac{3-\sqrt{5}}{3+\sqrt{5}} +\frac{3+\sqrt{5}}{3-\sqrt{5}}+\frac{12}{(3-\sqrt{5})(3+\sqrt{5})}]+3(3-\sqrt{5})(3+\sqrt{5})\\=\frac{(3-\sqrt{5})^{2}+(3+\sqrt{5})^{2}+12}{(3-\sqrt{5})(3+\sqrt{5})} +3(3^{2}-(\sqrt{5})^{2})\\=\frac{(3-\sqrt{5})^{2}+(3+\sqrt{5})^{2}+12}{(3^{2}-(\sqrt{5})^{2})} +12$

$=\frac{(3-\sqrt{5})^{2}+(3+\sqrt{5})^{2}+12}{4} +12\\=22$

Thus, the value of the expression is 22.