Question

if alfa and bita are the zeros of the polynomial 2x^2-7x+3. Evaluate alfa/bita+bita7alfa
​

Answer 1

Answer:

37 / 6

Step-by-step explanation:

2x^2 - 7x + 3 ⇒ 2[ x^2 - (7/2)x + (3/2) ]

 Poly. written in form of k( x^2 - Sx + P ) represent S and P as sum and product of their roots. Here, if α and β are roots:

        ⇒ α + β = (7/2)

        ⇒ αβ = (3/2)

⇒ α + β = (7/2)  ⇒ (α+β)^2 = (7/2)^2

⇒ α^2 + β^2 + 2αβ = 49/4

⇒ α^2 + β^2 = (49/4) - 2(3/2) { αβ = 3/2}

⇒ α^2 + β^2 = (49-12)/4 = 37/4

 

In question :

⇒ α/β + β/α

⇒ ( α^2 + β^2 ) / αβ

        α^2 + β^2 = 37/4 ; αβ = 3/2

⇒ ( 37/4 ) / ( 3/2 )

⇒  ( 37/4 ) * ( 2/3 )

⇒ 37 / 6

       Hence the required value is 37/6

Answer 2

$\Large\sf{\underline{\color{green}{CORRECT\:\:\: QUESTION:}}}$

$\small{\sf{If \alpha \& \beta \;are \;the\; roots\; of \;the\; polynomial\; 2x^2 - 7x + 3 . Then\; find\; \dfrac{\alpha}{\beta}+ \dfrac{\beta}{\alpha}}}$

$\Large\sf{\underline{\color{pink}{SOLUTION:}}}$

$\small \rm {We\; know\; that} \\\\ \small\sf{\alpha+\beta = \dfrac{-b}{a} \&\alpha\beta=\dfrac{c}{a}}\\\\\small\sf{\longrightarrow \frac{\alpha}{\beta}+\frac{\beta}{\alpha}= \frac{b^2 - 2ac}{ac}}\\\\\small\rm{\to b : coefficient \;of\; x= - 7}\\\\\small\rm{\to c:constant \; term=3}\\\\\to \small\rm{a :coefficient\quad of \quad x^2 = 2} \small\sf{\longrightarrow \dfrac{(-7)^2 - 2(2)(3)}{2(3)} = \boxed{\bf\frac{37}{6}}}$