Question

if a and b are the zeroes of the polynomial 5x²-7x+2then the sum of their reciprocal is​

Answer 1

Answer:

-7/2

Step-by-step explanation:

For a quadratic equation , x² + ax+b=0

with zeroes as x1,x2

a = x1 + x2

b=x1 × x2

So the equation can be written as

x²+(x1+x2)x+x1x2

Here, the equation is,

5x²-7x+2=0

x² - (7/5)x+2/5 = 0   (Dividing by 5)

Here let the zeroes be x1, x2

We know from the equation,

x1+ x2 = -7/5

x1×x2= 2/5

To find,

$\frac{1}{x1} + \frac{1}{x2}$

= $\frac{x1+x2}{x1 x2}$

=$\frac{\frac{-7}{5} }{\frac{2}{5} } = \frac{-7}{2}$

Answer 2

Solution :

We have quadratic polynomial p(x) = 5x² - 7x + 2

Zero of the polynomial p(x) = 0

$\longrightarrow\sf{5x^{2} - 7x + 2=0}\\\\\longrightarrow\sf{5x^{2} -5x -2x + 2=0}\\\\\longrightarrow\sf{5x(x-1 ) -2(x-1)=0}\\\\\longrightarrow\sf{(x-1)(5x-2)=0}\\\\\longrightarrow\sf{x-1=0\:\:\:Or\:\:\:5x-2=0}\\\\\longrightarrow\sf{x=1\:\:\:Or\:\:\:5x=2}\\\\\longrightarrow\bf{x=1\:\:\:Or\:\:\:x=2/5}$

∴ α = 1 & β = 2/5 are the zeroes of the polynomial .

$\underline{\boldsymbol{According\:to\:the\:question\::}}}$

Reciprocal of α = 1

Reciprocal of β = 5/2

$\mapsto\sf{\dfrac{1}{\alpha }+\dfrac{1}{\beta } =1+\dfrac{5}{2} }\\\\\\\mapsto\sf{\dfrac{1}{\alpha }+\dfrac{1}{\beta }=\dfrac{2+5}{2} }\\\\\\\mapsto\bf{ \dfrac{1}{\alpha }+\dfrac{1}{\beta }=\dfrac{7}{2} }$