Solve this all question.I will mark you brainliest .Answer it with solutions and explanations pls do it fast!! Solve only question no.4 all 3 sub questions
Answers
Answer:
1. Let the roots be a and b
a³ - b³ = 215 ........(1)
a − b = 5. ...............(2)
cubing both sides,
(a-b)³ = 5³
(a-b)³ = 125
a³ - b³ -3ab(a-b) = 125
215-3ab(5) = 125 [ From (1) & (2)]
3ab(5) = 215−125
15ab = 90
ab=6 .......................(3)
Now, (a +b)² = (a-b)² + 4ab
(a +b)² = 5² + 4×(6). [From. (2) & (3)]
(a +b)² = 25 + 24.
(a +b)² = 49
a+b = ±7...............(4)
Standard form of the equation is:
Standard form of the equation is: x²- (a+b)x + ab = 0
From (3) and (4)
x² ±7x + 6 =0
2.Let the fixed charge be Rs. x
and the extra charge be Rs. y
x + 4y = 27.........................(1)
x+ 2y = 21 ........................(2)
On substracting (1) and (2)
2y = 6
y = 3
putting y in (1)
x + 4(3) = 27
x = 27 - 12
x = 15
∴x = Rs.15 and y = Rs.3
∴ Fixed charge =Rs. 15
∴ Charge for each extra day =Rs. 3
3. According to the questions
AP is 1600, 2100, 2600.....................
a =1600 , d=500 , Sn = 38500
Sn = n/2[2a+(n-1)d]
38500 = n/2 [2*1600 +(n - 1) 500 ]
77000 = n[ 3200 +500n - 500]
77000= 2700n +500n^2
5n² + 27n - 770 =0
5n² -50n+77n -770 = 0
5n(n-10) + 77(n-10) = 0
(5n+77) (n-10) = 0
n = -77/5 (reject) and. n = 10
n = 10
Therefore, In 10 years he saves 38,500.
Hope you satisfied with my answer.
Please mark as brainly.