If a and b are the 2 zeroes of the quadratic polynomial x^2-3x 7 , find a quadratic polynomial whose zeroes are 1/a & 1/b?

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Answer

Answer

P(x)=x²-3x+7,
sum of zeros of a quadratic polynomial=a+b,
a+b=-(-3)/1=3---(1).
product of zeros=a.b,
a.b=c/a=7/1=7--(2).
to find-->f(x) having zeroes 1/a and 1/b ,
sum of zeroes of f(x)=1/a+1/b=a+b/ab=3/7,
product of zeros of f(x)=1/a×1/b=1/ab=1/7,
f(x)=x²-( sum of zeros)x+ product of zeros,
=x²-3/7x+1/7,
=7x²-3x+1

Answer

And 1/b.

Since a and b are the zeroes of the polynomial x^2-3x+7, we can write:

x^2 - 3x + 7 = (x - a)(x - b)

Expanding the right-hand side, we get:

x^2 - (a + b)x + ab = x^2 - 3x + 7

Equating the coefficients of x and the constant terms, we get:

a + b = 3
ab = 7

We want to find a quadratic polynomial whose zeroes are 1/a and 1/b. So, let's write:

(x - 1/a)(x - 1/b) = 0

Expanding the right-hand side, we get:

x^2 - (1/a + 1/b)x + 1/(ab) = 0

We can simplify 1/(ab) as follows:

1/(ab) = 1/7

Substituting the values of 1/a, 1/b, and 1/(ab) in the above equation, we get:

x^2 - [(b + a)/(ab)]x + 1/7 = 0

Using the values of a + b and ab that we obtained earlier, we can simplify the above equation as follows:

x^2 - (3/7)x + 1/7 = 0

Therefore, the required quadratic polynomial whose zeroes are 1/a and 1/b is:

x^2 - (3/7)x + 1/7

Answer

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