1.

Find the zeroes of the quadratic polynomial (8x2 ˗ 4) and verify the relation between the zeroes and the coefficients.

Answer»

We have: 

f(x) = 8x2 ˗ 4 

It can be written as 8x2 + 0x ˗ 4 

= 4 { (√2x)2 ˗ (1)2 } 

= 4 (√2x + 1) (√2x ˗ 1) 

∴ f(x) = 0 ⇒ (√2x + 1) (√2x ˗ 1) = 0 

⇒ (√2x + 1) = 0 or √2x ˗ 1 = 0

⇒ x = \(\frac{-1}{\sqrt2}\) or x = \(\frac{1}{\sqrt2}\)

So, the zeroes of f(x) are \(\frac{-1}{\sqrt2}\) and \(\frac{1}{\sqrt2}\)

Here the coefficient of x is 0 and the coefficient of x2 is √2

Sum of zeroes = \(\frac{-1}{\sqrt2}\) + \(\frac{1}{\sqrt2}\) = \(\frac{-1+1}{\sqrt2}\) = \(\frac{0}{\sqrt2}\) = \(\frac{-(coefficient\,of\,x)}{(coefficient\,of\,x^2)}\)

Product of zeroes = \(\frac{-1}{\sqrt2}\) x \(\frac{1}{\sqrt2}\) = \(\frac{-1\times4}{2\times4}\) = \(\frac{-4}{8}\) = \( \frac{constant\,term}{(coefficient\,of\,x^2)}\)


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