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kramer
3 years ago
6

Find two consecutive numbers whose squares differ by 33

Mathematics
1 answer:
stich3 [128]3 years ago
3 0

Answer:

Two consecutive numbers whose squares differ by 33 are 16 and 17

Step-by-step explanation:

lets assume first number be x

since numbers are consecutive , so other number will be x + 1

From given information in question

(x + 1)² - x² = 33

⇒ (x² + 1² + 2x)  - x² = 33       [   since (a+b)² = a² + b² + 2ab ]

⇒ x² + 1² + 2x - x² = 33

⇒ 2x + 1 = 33

⇒ 2x = 33 - 1

⇒ x = 32/2 = 16

so one number is x = 16 and other number is x + 1 = 16 + 1 = 17

lets recheck our solution

17² - 16² = 289 - 256 = 33 , And since difference is 33 , two required consecutive numbers are 16 and 17.


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The error made by zacharias in solving the quadratic equation 0 = -2x² + 5x - 3 using quadratic formula is; the 2 in the numerator should be –2.

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There are four methods of solving quadratic equation. Namely;

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x = -b ± √b² - 4ac / 2a

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Therefore, the solution to the quadratic equation is x = -3 or -2

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