Question

C. 8
d. 16
. Find x such that:
. (7/4) 3 * (7/4).5 = (7/4)x-2
. (125/8) * (125/8)* = (5/2)18
Simplify:​

Answer 1

Answer:

x² + (x + 1)2 = 85

⇒ x² + x² + 2x + 1 = 85

⇒ 2x² + 2x + 1 – 85 = 0

⇒ 2x² + 2x – 84 = 0

⇒ x² + x – 42 = 0

⇒ x² + 7x – 6x – 42 = 0

⇒ x (x + 7) – 6 (x + 7) = 0

⇒ (x + 7) (x – 6) = 0

Either x + 7 = 0, then x = -7 or x – 6 = 0, then x = 6

(i) If x = -7, then the first number = -7 and second number = -7 + 1 = -6

(ii) If x = 6, then the first number = 6 and second number = 6 + 1 = 7

Hence numbers are -7, -6 or 6, 7

Question 2.

Divide 29 into two parts so that the sum of the squares of the parts is 425.

Solution:

Total = 29

Let first part = x

Then second part = 29 – x

According to the condition

x² + (29 – x)2 = 425

⇒ x² + 841 + x² – 58x = 425

⇒ 2x² – 58x + 841 – 425 = 0

⇒ 2x² – 58x + 416 = 0

⇒ x² – 29x + 208 = 0 (Dividing by 2)

⇒ x² – 13x – 16x + 208 = 0

⇒ x(x – 13) – 16 (x – 13) = 0

⇒ (x – 13) (x – 16) = 0

Either x – 13 = 0, then x = 13 or x – 16 = 0, then x = 16

(i) If x = 13, then First part =13 and second part = 29 – 13 = 16

(ii) If x = 16, then First part =16 and second part = 29 – 16 = 13

Parts are 13, 16

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