Question

Example 2 : Find the zeroes of the quadratic polynomial x^2+ 7x + 10, and verify the relationship between the zeroes and the coefficients.​

Answer 1

Answer:

factorize it by middle term splitting

$x {}^{2} + 7x + 10 \\x {}^{2} + 5x + 2x + 10 \\ x(x + 5) + 2(x + 5)$

it means -2 &-5 are zeroes sum of zeroes -5-2=-7

and product -5×-2=10

here a=1,b=7 and c=19

to verify their relation sum -b/a = -5/1= -5

and product c/a=10/1=10

and hence relation is verified

........hope it will be helpful

Answer 2

$\bf ➯p(x) = {x}^{2} + 7x + 10$

$\bf ➣let$

$\bf ⇒{x}^{2} + 7x + 10 = 0$

$\bf ⇒{x}^{2} + 2x + 5x + 10 = 0$

$\bf ⇒x(x + 2) + 5(x + 2) = 0$

$\bf➾(x + 5)(x + 2) = 0$

$➢So$

$\bf ➾x = ( - 2) \: and \: ( - 5)$

$\bf ➣Therefore$

$\bf ⇒a = - 2 \: and \: b = - 5$

$\bf ➯sum \: of \: zeros \: = \dfrac{coefficient \: of \: x}{coefficient \: of \: {x}^{2} }$

$\bf ⇒a + b = \dfrac{ - b}{a}$

$\bf ⇒- 2 + ( - 5) = \dfrac{ - 7}{1}$

$\bf ⇒- 7 = - 7$

$\bf ➯product \: of \: zeroes = \dfrac{c}{d}$

$\bf ⇒a \times b = \dfrac{c}{d}$

$\bf⇒( - 2)( - 5) = \dfrac{10}{1}$

$\bf ➾10 = 10$

Since,

L.H.S = R.H.S

Hope it helps you...