Solve this question? (grade 10) find "a" so that the distance between points (-2,-3) and (-3, a) is equal to 5
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Solve this question? (grade 10) find "a" so that the distance between points (-2,-3) and (-3, a) is equal to 5

[From: ] [author: ] [Date: 12-03-11] [Hit: ]
now use X = (-B + or - sqrt (B^2 - 4AC)) / 2AAnd you will get their answers.-√[ (-3 -(-2))² + (a-(-3))² ]=5√[ (-3 -(-2))² + (a-(-3))² ] ²=5 ²√[ (- 1)² + (a + 3)² ] ²=5 ²√[ 1 + (a + 3)² ] ²=5 ²1 + (a + 3)² = 25(a + 3)² = 24√(a + 3)² = + - √24=+ - 2√6a + 3 = + - 2√6a = - 3 + - 2√6...........
can you please simply explain how to solve this? I though it would be 5= √(-3 -(-2))² + (a-(-3))² but

apparently I'm wrong. The answer's supposed to be either a= -3 + 2√6 or a= -3- 2√6

anyways, I'd appreciate it if you could explain it in simple terms please
thanks!

-
Your orginal solution is right, you just need to solve for the unknown "a".
First expand your orginal solution and square both sides and get a^2 + 6a +10 = 25
which equals a^2 + 6a - 15 which has the form AX^2 + BX +C where A, B, and C are coefficients.

now use
X = (-B + or - sqrt (B^2 - 4AC)) / 2A

And you will get their answers.

-
√[ (-3 -(-2))² + (a-(-3))² ] = 5

√[ (-3 -(-2))² + (a-(-3))² ] ² = 5 ²

√[ (- 1)² + (a + 3)² ] ² = 5 ²

√[ 1 + (a + 3)² ] ² = 5 ²

1 + (a + 3)² = 25

(a + 3)² = 24

√(a + 3)² = + - √24 = + - 2√6

a + 3 = + - 2√6

a = - 3 + - 2√6.............ANSWERS

Mathcerely, Robert Jones "Teacher/Tutor of Fine Students"
Moved Florida to France June 2011 but still tutor world-wide (USA , UK ,
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