Question

find the zeros of the polynomial x2+ 13x + 36 and verify the relation between the zeroes and the coefficients

Answer 1

Step-by-step explanation:

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Answer 2

Step-by-step explanation:

${x}^{2}$ + 13x + 36

${x}^{2}$ + (9x + 4x) + 36

x (x + 9) + 4 (x + 9)

(x + 4) (x + 9)

• x + 4 = 0
• x = -4

• x + 9 = 0
• x = -9

let ,

a = 1

b = 13

$\alpha$ = -4

$\beta$ = -9

Sum of zeros = $\alpha$ + $\beta$ = $\frac{-b}{a}$

=> -4 + (-9) = $\frac{-13}{1}$

=> -4 - 9 = -13

=> -13 = -13

:. $\alpha$ + $\beta$ = $\frac{-b}{a}$

let

a = 1

c = 36

Sum of products = $\alpha$ × $\beta$ = $\frac{c}{a}$

=> -4 × -9 = $\frac{36}{1}$

=> 36 = 36

:. $\alpha$ × $\beta$ = $\frac{c}{a}$

Hence proved.