Question

39a²-33a-6=0 in quadratis equation pls find the answer

Answer 1

Answer:

Given, a + b + ab = 8 )

b + c + bc = 15 )

c + a + ca = 35 )

we have to find (a + b + c + abc) = ?

from equation (ii),

b + c + bc = 15

b + c(b + 1) = 15

c = (15 - b)/(b + 1)

put it in equation (iii),

(15 - b)/(b + 1) + a + (15 - b)a/(b+ 1) = 35

(15 - b) + a(b + 1) + (15 - b)a = 35(b + 1)

15 - b + ab + a + 15a - ab = 35b + 35

15 + a - b + 15a = 35b + 35

a - b + 15a - 35b = 35 - 15 = 20

16a - 36b = 20

4a - 9b = 5

4a = 5 + 9b

a = (5 + 9b)/4 )

put equation (iv) in equation (i),

(5 + 9b)/4 + b + (5 + 9b)/4 × b = 8

(5 + 9b) + 4b + 5b + 9b² =8 × 4

18b + 9b² + 5 = 32

9b² + 18b - 27= 0

b² + 2b - 3 = 0

(b + 3)(b - 1) = 0.

b = -3 and 1 but a, b , c are positive numbers

so, b = 1

put equation (iv), a = (5 + 9)/4 = 7/2

c = (15 - b)/(b + 1) = (15 - 1)/(1 + 1) = 7

now, a + b + c + abc = 7/2 + 1 + 7 + 7/2 × 1 × 7

= 7/2 + 8 + 49/2

= 56/2 + 8

=28 + 8 = 36

hence, answer is 36

Answer 2

$\huge\red{Answer}$

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☆☞ [ Verified answer ]☜☆

$Answer:</p><p></p><p>a=−213 or a=1</p><p></p><p>$

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