More quadratics

[Mac_dw]

Could you please check my work? See if I'm cetchin on

X^2 - 9X - 4 = 6

X^2 -9x - 10 = 0

X = { -(-9) +/- sqrt[(-9)^2 - 4(1)(-10) } / 2

X = { 9 +/- sqrt[81 + 40] } / 2

X = {9 +/- 11} / 2

X= 10, X= (-1)

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6X^2 - 23x + 10 = 0

X = { -(-23) +/- sqrt[(-23)^2 - 4(6)(10) } / 12

X= { 23 +/- sqrt[529 - 240] } / 12

X = {23 +/- sqrt289} / 12 (is the final solution or does it reduce down?)

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X^2 + 2X - 5 = 0

X = { -(2) +/- sqrt[2^2 -4(1)(-5) } / 2

X = { -2 +/- sqrt[24] } / 2

X = { (-2) +/- (2 sqrt6) } / 2

X = -1 +/- sqrt6

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[tkhunny]

17*17 = 289

If you check your answers, you can know for yourself that they are correct.

 

[Mac_dw]

Yeah I see your point looking back at it. These are problems I had previously missed, guess my confidence was down. Not sure I could check my work with the last one, however, I got it wrong when i did???

Thank You. I appreciate your help. And the help of all involved. I feel a lot stronger heading into my final later this week.

VIVA La Algebra!!!

 

[Denis]

X^2 - 9X - 4 = 6
X^2 -9x - 10 = 0
X = { -(-9) +/- sqrt[(-9)^2 - 4(1)(-10) } / 2
X = { 9 +/- sqrt[81 + 40] } / 2
X = {9 +/- 11} / 2
X= 10, X= (-1)

Correcto!
You could have solved this by factoring:
x^2 -9x - 10 = 0
(x - 10)(x + 1) = 0
x = 10 or x = -1
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6X^2 - 23x + 10 = 0
X = { -(-23) +/- sqrt[(-23)^2 - 4(6)(10) } / 12
X= { 23 +/- sqrt[529 - 240] } / 12
X = {23 +/- sqrt289} / 12 (is the final solution or does it reduce down?)

C'mon Mac: sqrt(289) = 17 :idea:
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X^2 + 2X - 5 = 0
X = { -(2) +/- sqrt[2^2 -4(1)(-5) } / 2
X = { -2 +/- sqrt[24] } / 2
X = { (-2) +/- (2 sqrt6) } / 2
X = -1 +/- sqrt6

Yepper :!:
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