Math, asked by lakshmic064, 9 months ago

find the zeroes of the polynomial x²+7x+10 and verify the relationship between the zeroes and the coefficient​

Answers

Answered by Himanidaga
9

Step-by-step explanation:

 {x}^{2}  + 7x + 10

 {x}^{2}   + 5x + 2x + 10

x ( x + 5) + 2( x + 5)

(x + 2) ( x + 5)

x = -2 and x = -5

alfa + Bita = -b/a < => alfa × Bita = c/a

-2-5 = -7/1 -2 × -5 = 10

Hence, Verified

Answered by Anonymous
42

Answer:

Given:

  • p(x) = x² + 7x + 10

To do:

  • Find the zeroes
  • Verification

\sf{\underline{Solution:}}

We can find the zeroes by simply factorising the polynomial

So factorising we get:

= x² + 7x + 10 = 0

= x² + 5x + 2x + 10 = 0

= x (x + 5) + 2 (x + 5) = 0

= (x + 5)(x + 2) = 0

= x + 5 = 0⠀⠀⠀⠀⠀⠀or⠀⠀⠀⠀⠀x + 2 = 0

= ⠀x = -5⠀⠀⠀⠀⠀⠀⠀or⠀⠀⠀⠀⠀⠀x = -2

So the zeroes are -5 , -2

Now verification:

For sum of zeroes , the relation used:

\boxed{\sf{Sum \ of \ zeroes = \dfrac{- Coefficient \ of \ x}{Coefficient \ of \ x^2}}}

Putting the values:

- 5 +(-2) = \dfrac{-7}{1}

-7 = -7

For product of zeroes , the relation used:

\boxed{\sf{Product \ of \ zeroes = \dfrac{Constant \ term}{Coefficient \ of \ x^2}}}

Putting all the values:

-5 \times -2 = \dfrac{10}{1}

10 = 10

_____________________________

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