Question

Pls answer only 16 and 19

Answer 1

Question 16:

$\alpha$ and $\beta$ are the roots of the equation.

That means $(x-\alpha )(x-\beta ) = 2x^{2} -6x+5 = 0$

$(x-\alpha )(x-\beta ) = 2x^{2} -6x+5 = 0$

          $2x^{2} -6x+5 = 0\\x^{2} -3x+\frac{5}{2} =0$

          $(x-\alpha )(x-\beta ) =x^{2} - (\alpha +\beta )x +\alpha \beta$

So,

$x^{2} - (\alpha +\beta )x +\alpha \beta = x^{2} -3x+\frac{5}{2}$

Compare coefficient

$\alpha +\beta = 3\\\alpha \beta = \frac{5}{2}$

Consider $\frac{\beta }{\alpha } + \frac{\alpha }{\beta }$

$\frac{\beta }{\alpha } + \frac{\alpha }{\beta } =\frac{ \alpha ^{2} +\beta ^{2}}{\alpha \beta }$

$= \frac{(\alpha +\beta )^{2} -2\alpha \beta }{\alpha \beta } \\= \frac{3^{2} -2*\frac{5}{2} }{\frac{5}{2} } \\= 4*\frac{2}{5} \\=\frac{8}{5}$

Ans: 8/5

Question 19:

$f(x+2) = 2x^{2} +5x-3$

Substitute x with -1

So,

$f(1) = 2(-1)^{2} +5(-1)-3\\=2-5-3\\=-6$

Ans: -6