Find the zeroes of the quadratic polynomial 2x2 ˗ 11x + 15 and verify

1.	Find the zeroes of the quadratic polynomial 2x2 ˗ 11x + 15 and verify the relation between the zeroes and the coefficients.
Answer» f(x) = 2x² ˗ 11x + 15 = 2x² ˗ (6x + 5x) + 15 = 2x² ˗ 6x ˗ 5x + 15 = 2x (x ˗ 3) ˗ 5 (x ˗ 3) = (2x ˗ 5) (x ˗ 3) ∴ f(x) = 0 ⇒ (2x ˗ 5) (x ˗ 3) = 0 ⇒ 2x ˗ 5= 0 or x ˗ 3 = 0 ⇒ x = \(\frac{5}2\) or x = 3 So, the zeroes of f(x) are \(\frac{5}2\) and 3. Sum of zeroes = \(\frac{5}2\) + 3 = \(\frac{5+6}2\) = \(\frac{11}2\) = \(\frac{-(coefficient\,of\,x)}{(coefficient\,of\,x^2)}\) Product of zeroes = \(\frac{5}2\) x 3 = \(\frac{-15}2\) = \(\frac{constant\,term}{(coefficient\,of\,x^2)}\)

Find the zeroes of the quadratic polynomial 2x2 ˗ 11x + 15 and verify the relation between the zeroes and the coefficients.

Answer»

f(x) = 2x² ˗ 11x + 15

= 2x² ˗ (6x + 5x) + 15

= 2x² ˗ 6x ˗ 5x + 15

= 2x (x ˗ 3) ˗ 5 (x ˗ 3)

= (2x ˗ 5) (x ˗ 3)

∴ f(x) = 0 ⇒ (2x ˗ 5) (x ˗ 3) = 0

⇒ 2x ˗ 5= 0 or x ˗ 3 = 0

⇒ x = \(\frac{5}2\) or x = 3

So, the zeroes of f(x) are \(\frac{5}2\) and 3.

Sum of zeroes = \(\frac{5}2\) + 3 = \(\frac{5+6}2\) = \(\frac{11}2\) = \(\frac{-(coefficient\,of\,x)}{(coefficient\,of\,x^2)}\)

Product of zeroes = \(\frac{5}2\) x 3 = \(\frac{-15}2\) = \(\frac{constant\,term}{(coefficient\,of\,x^2)}\)