Question

If a & b are the zeroes of polynomial p(x) = x2 + x - 1, then find 1/a + 1/b​

Answer 1

Answer:

Step-by-step explanation:

f (x) = x^2 + x + 1

Where, a = 1 , b = 1 , c =1

One zero = a

Second zero = b

Sum of zeroes ( a + b ) = - b/a = -1 / 1 = -1

Product of zeroes ( a × b ) = c/a = 1 / 1 = 1

We have to find value of , 1 / a + 1 / b

Taking LCM

= a + b / ab

Keeping the values now,

= -1 / 1

= -1

Thus, 1 / a + 1 / b = -1

________________

Hope it helps...!!!

another method

According to root's or zeroes concept of quadratic equation..

If a and b are zeroes of polynomial x^2 + x+1 (form of px^2 + qx+c) then..

Since A/Q..

p=1, q= 1 and c = 1.

=) a+b = - q/p

=) a+b = - 1/1 = - 1

And

=) a*b = c/p

=) a*b = 1/1 = 1

----------------------------

To find : 1/a + 1/b :

Take lcm of a and b = ab

= (b+a) /ab

Put value of a+b and ab.

= - 1/1

= - 1

Hope it helps u :)

follow me guys plz plz

Answer 2

Answer:

$\sf{\frac{1}{a}+\frac{1}{b}=1}$

Step-by-step explanation:

               $\sf{Given\:p(x)=x^{2} +x-1$

$\sf{Sum\:of\:zeroes(a+b)=\frac{-b}{a}$

                     $\sf{(a+b)=\frac{-1}{1}}$

                     $\sf{(a+b)=-1}$

$\sf{Product\:of\:zeroes(ab)=\frac{c}{a}$

                         $\sf{(ab)=\frac{-1}{1}$

                         $\sf({ab)=-1$

$\frac{1}{a}+\frac{1}{b}=\frac{a+b}{ab}$

$\frac{1}{a}+\frac{1}{b}=\frac{-1}{-1}$

$\frac{1}{a}+\frac{1}{b}=1$

$\bf\pink{\underline{Please\:mark\:it\:as\:brainlist\:answer}$