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igomit [66]
3 years ago
7

What number must be added to x^2+6x+7 to form a perfect square

Mathematics
1 answer:
RSB [31]3 years ago
7 0
Let the number be k,

Then adding to the expression we have,

{x}^{2}  + 6x + 7 + k
For this to be a perfect square, the discriminant must be zero.

We have,

a=1, b=6, c=7+k


{b}^{2}  - 4ac = 0

This implies that

{6}^{2}  - 4(1)(7 +k ) = 0
36 - 28 - 4k = 0
8 - 4k = 0
8 = 4k
k = 2
hence the number to be added is 2
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dlinn [17]
ANSWER

The solution is
(-4,-1)


EXPLANATION

The given equations are:

2x + 5y =  - 13...(1)

and

3x - 4y =  - 8...(2)


We multiply equation 1 by 3 to get,

6x + 15y =  - 39...(3)

We multiply equation 2 also by 2 to get,

6x - 8y =  - 16...(4)

We subtract equation 4 from equation 3 to get,


23y =  - 23
This implies that,

y =  - 1

We put this value of y into any of the equations say equation 2

3x - 4( - 1) =  - 8
Simplify

3x + 4 =  - 8
Group like terms
3x =  - 8 - 4
Simplify,

3x =  - 12

Divide both sides by 3 to get,

x =  - 4


Therefore the solution is
(-4,-1)

The correct answer is option D.
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