Question 16:
[tex]\alpha[/tex] and [tex]\beta[/tex] are the roots of the equation.
That means [tex](x-\alpha )(x-\beta ) = 2x^{2} -6x+5 = 0\\[/tex]
[tex](x-\alpha )(x-\beta ) = 2x^{2} -6x+5 = 0[/tex]
[tex]2x^{2} -6x+5 = 0\\x^{2} -3x+\frac{5}{2} =0[/tex]
[tex](x-\alpha )(x-\beta ) =x^{2} - (\alpha +\beta )x +\alpha \beta[/tex]
So,
[tex]x^{2} - (\alpha +\beta )x +\alpha \beta = x^{2} -3x+\frac{5}{2}[/tex]
Compare coefficient
[tex]\alpha +\beta = 3\\\alpha \beta = \frac{5}{2}[/tex]
Consider [tex]\frac{\beta }{\alpha } + \frac{\alpha }{\beta }[/tex]
[tex]\frac{\beta }{\alpha } + \frac{\alpha }{\beta } =\frac{ \alpha ^{2} +\beta ^{2}}{\alpha \beta }[/tex]
[tex]= \frac{(\alpha +\beta )^{2} -2\alpha \beta }{\alpha \beta } \\= \frac{3^{2} -2*\frac{5}{2} }{\frac{5}{2} } \\= 4*\frac{2}{5} \\=\frac{8}{5}[/tex]
Ans: 8/5
Question 19:
[tex]f(x+2) = 2x^{2} +5x-3\\[/tex]
Substitute x with -1
So,
[tex]f(1) = 2(-1)^{2} +5(-1)-3\\=2-5-3\\=-6[/tex]
Ans: -6
