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gtnhenbr [62]
3 years ago
7

For what value of $m$ does the equation $(x+4)(x+1) = m + 2x$ have exactly one real solution? express your answer as a common fr

action.
Mathematics
1 answer:
Vladimir [108]3 years ago
5 0

<span>(x+4)(x+1) = m+2x
x² + 5x + 4 − m − 2x = 0
x² + 3x + (4−m) = 0

Quadratic equation has exactly one real solution
when discriminant = 0

b² − 4ac = 0
3² − 4(1)(4−m) = 0
9 − 16 + 4m = 0
−7 + 4m = 0
4m = 7
m = 7/4

-------

Second problem:

Line y = 3 intersects graph of y = 4x² + x − 1 at point A and B

y-coordinates of points A and B = 3, since they are both on line y = 3. We need to find x-coordinates

4x² + x − 1 = 3
4x² + x − 4 = 0

x = (−1 ± √(1−4(4)(−4))) / 8
x = (−1±√65) / 8

AB = (−1+√65)/8 − (−1−√65)/8 = 2√65/8 = √65/4 = √(65/16)

m−n = 49</span>

I have take the answer from internet, I hope it help


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5 -2x + 6y= -38<br> ? 3x – 4y = 32<br> •(-4, - 5)<br> •(-5, 4)<br> •(1, – 6)<br> (4, - 5)
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Answer:

Option D (4, -5)

Step-by-step explanation:

This question can be solved by various methods. I will be using the hit and trial method. I will plug in all the options in the both the given equations and see if they  balance simultaneously.

Checking Option 1 by plugging (-4, -5) in the first equation:

-2(-4) + 6(-5) = -38 implies 8 - 30 = -38 (not true).

Checking Option 2 by plugging (-5, 4) in the first equation:

-2(-5) + 6(4) = -38 implies 10 + 24 = -38 (not true).

Checking Option 3 by plugging (1, -6) in the second equation:

3(1) - 4(-6) = 32 implies 3 + 24 = 32 (not true).

Since all the options except Option 4 have been ruled out, therefore, (4,-5) is the correct answer!!!

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I believe your answer is D. 77. 
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