Question

if α and β are the roots of 4x^2-9x+16=0 find √α +√β​

Answer 1

if α and β are the roots of 4x^2-9x+16=0 then √α +√β​  = $\dfrac{5}{2}$

Step-by-step explanation:

Given the polynomial $4\textrm{x}^{2} -9\textem{x}+16$

which is similar to the polynomial $\textrm{a}\textrm{x}^{2} +\textrm{b}\textrm{x}+\textrm{c}$

$\textrm{a}=4,\textrm{b}=-9,\textrm{c}=16$

Let $\alpha ,\beta$ be the roots of the polynomial

$\alpha +\beta =-\dfrac{\textrm{b}}{\textrm{a}} =-\dfrac{-9}{4} =\dfrac{9}{4}$

$\alpha \beta =\dfrac{\textrm{c}}{\textrm{a}} =\dfrac{16}{4} =4$

$(\sqrt{\alpha }+\sqrt{\beta })^{2}=\left ( \sqrt{\alpha } \right )^{2}+\left ( \sqrt{\beta } \right )^{2}+2\sqrt{\alpha }\sqrt{\beta }$

 $\Rightarrow (\sqrt{\alpha }+\sqrt{\beta })^{2}=\left ( \sqrt{\alpha } \right )^{2}+\left ( \sqrt{\beta } \right )^{2}+2\sqrt{\alpha\beta }$

$\Rightarrow (\sqrt{\alpha }+\sqrt{\beta })^{2}= {\alpha } + {\beta } +2\sqrt{\alpha\beta }$

$\Rightarrow (\sqrt{\alpha }+\sqrt{\beta })^{2}= \dfrac{9}{4}+ 2\sqrt{4}$

$\Rightarrow (\sqrt{\alpha }+\sqrt{\beta })^{2}= \dfrac{9}{4}+ 2\times2$

$\Rightarrow (\sqrt{\alpha }+\sqrt{\beta })^{2}= \dfrac{9}{4}+ 4$      

$\Rightarrow (\sqrt{\alpha }+\sqrt{\beta })^{2}= \dfrac{9+16}{4}$

$\Rightarrow (\sqrt{\alpha }+\sqrt{\beta })^{2}= \dfrac{25}{4}$

$\Rightarrow (\sqrt{\alpha }+\sqrt{\beta })= \sqrt{\dfrac{25}{4}}$

$\Rightarrow (\sqrt{\alpha }+\sqrt{\beta })= \dfrac{5}{2}$