Question

if alpha and beta are the zeroes of polynomial 2x^2-7x+3.find the value of 1/alpha+1/beta​

Answer 1

Solution :

We have quadratic polynomial p(x) = 2x² - 7x + 3 & as zero of the polynomial p(x) = 0;

$\longrightarrow\sf{2x^{2} - 7x + 3=0}\\\\\longrightarrow\sf{2x^{2} - 6x - x +3=0}\\\\\longrightarrow\sf{2x(x-3) -1(x-3)=0}\\\\\longrightarrow\sf{(x-3)(2x-1)=0}\\\\\longrightarrow\sf{x-3=0\:\:\:Or\:\:\:2x-1=0}\\\\\longrightarrow\sf{x=3\:\:\:Or\:\:\:2x=1}\\\\\longrightarrow\bf{x=3\:\:\:Or\:\:\:x=1/2}$

∴ α = 3 & β = 1/2 are the two zeroes of the given polynomial.

As we know that given polynomial compared with ax² + bx + c;

• a = 2
• b = -7
• c = 3

$\underline{\mathcal{SUM\:OF\:THE\:ZEROES\::}}$

$\mapsto\sf{\alpha +\beta =\dfrac{-b}{a} =\bigg\lgroup \dfrac{Coefficient\:of\:x}{Coefficient\:of\:x^{2}} \bigg\rgroup }\\\\\\\mapsto\sf{3 + \dfrac{1}{2} }\\\\\\\mapsto\sf{\dfrac{6+1}{2} }\\\\\\\mapsto\bf{\dfrac{7}{2} }$

$\underline{\mathcal{PRODUCT\:OF\:THE\:ZEROES\::}}$

$\mapsto\sf{\alpha\times \beta =\dfrac{c}{a} =\bigg\lgroup \dfrac{Constant\:term}{Coefficient\:of\:x^{2}} \bigg\rgroup }\\\\\\\mapsto\sf{3 \times \dfrac{1}{2} }\\\\\\\mapsto\bf{\dfrac{3}{2}}$

Now;

$\longrightarrow\sf{\dfrac{1}{\alpha } +\dfrac{1}{\beta } }\\\\\\\longrightarrow\sf{\dfrac{\beta +\alpha }{\alpha \beta } }\\\\\\\longrightarrow\sf{\dfrac{\alpha +\beta }{\alpha \beta }}\\\\\\\longrightarrow\sf{\dfrac{7/2}{3/2} }\\\\\\\longrightarrow\sf{\dfrac{7}{\cancel{2}} \times \dfrac{\cancel{2}}{3} }\\\\\\\longrightarrow\boxed{\bf{\dfrac{7}{3} }}}$

Thus;

The value of 1/α + 1/β will be 7/3 .

Answer 2

According to the data in the question

We have to find the zero of polynomial and the value of expression

$\frac{1 }{ \alpha } + \frac{1}{ \beta }$

Step-by-step explanation:

Given equation is 2x²-7x+3

STEP-1:

By using splitting method :

The general equation is ax²+bx+c

Given equation is 2x²-7x+3

Now compare equation and general equation,

a=2 ,b=-7 and c=3

STEP-2:

a×c = 6

b= -6×-1

STEP-3:

$2x²-7x+3 = 0$

Split the mid value ,

$2x²-6x - x+3 = 0$

$(2x²-6x) -( x - 3) = 0$

Take out the 2x common from the 1st term,

$2x(x - 3) - 1(x - 3) = 0$

Take out (x-3) Common

$(x - 3)(2x - 1) = 0$

The values of x are ,

$x = 3 \: and \: \frac{1}{2}$

Here ,

$\alpha = 3$

$\beta = \frac{1}{2}$

STEP-4:

$\frac{1 }{ \alpha } + \frac{1}{ \beta }$

Put the values , we get

$\frac{1 }{ 3 } + \frac{1}{ \frac{1}{2} }$

Or

$\frac{1 }{ 3 } + \frac{2}{ 1 }$

Take LCM of 3 and 1 , we get the 3

$\frac{1 + 6}{3}$

Add the values ,

$\frac{7}{3}$

Hence ,

$\frac{1 }{ \alpha } + \frac{1}{ \beta } =\frac{1 }{ 3 } + \frac{1}{ \frac{1}{2} } = \frac{7}{3}$

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