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hichkok12 [17]
3 years ago
13

Ten times an integer is added to seven times it’s square. If the result is 152, what was the original number?

Mathematics
1 answer:
Bess [88]3 years ago
3 0

Answer:

Required number is 4.

Step-by-step explanation:

Let the required number be a.

Given,

Sum of ten times the integer and seven times it’s square is 152.

= > Ten times of a + seven times of it's square = 152

= > 10( a ) + 7( a )^2 = 152

= > 10a + 7a^2 - 152 = 0

= > 7a^2 + 10a - 152 = 0

= > 7a^2 + ( 38 - 28 )a - 152 = 0

= > 7a^2 + 38a - 28a - 152 = 0

= > a( 7a + 38 ) - 4( 7a + 38 ) = 0

= > ( a - 4 )( 7a + 38 ) = 0

= > a = 4 or - 38 / 7

Hence the required number is 4.

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(i) Write the expansion of (x + y)² and (x - y)². (ii) Find (x + y)² - (x - y)² (iii) Write 12 as the difference of two perfect
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Answer:

1a (x + y)² = x² + 2xy + y²

1b. (x - y)² = x² - 2xy + y²

2. (x + y)² - (x - y)² = 4xy

3. 4² – 2² = 12

Step-by-step explanation:

1a. Expansion of (x + y)²

(x + y)² = (x + y)(x + y)

(x + y)² = x(x + y) + y(x + y)

(x + y)² = x² + xy + xy + y²

(x + y)² = x² + 2xy + y²

1b. Expansion of (x - y)²

(x - y)² = (x - y)(x - y)

(x - y)² = x(x - y) - y(x - y)

(x - y)² = x² - xy - xy + y²

(x - y)² = x² - 2xy + y²

2. Determination of (x + y)² - (x - y)²

This can be obtained as follow

(x + y)² = x² + 2xy + y²

(x - y)² = x² - 2xy + y²

(x + y)² - (x - y)² = x² + 2xy + y² - (x² - 2xy + y²)

= x² + 2xy + y² - x² + 2xy - y²

= x² - x² + 2xy + 2xy + y² - y²

= 2xy + 2xy

= 4xy

(x + y)² - (x - y)² = 4xy

3. Writing 12 as the difference of two perfect square.

To do this, we shall subtract 12 from a perfect square to obtain a number which has a perfect square root.

We'll begin by 4

4² – 12

16 – 12 = 4

Find the square root of 4

√4 = 2

4 has a square root of 2.

Thus,

4² – 12 = 4

4² – 12 = 2²

Rearrange

4² – 2² = 12

Therefore, 12 as a difference of two perfect square is 4² – 2²

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