標題:
Quadratic Equation (Urgent)
發問:
Please list out the steps and explain why clearly. Thanks!You don't necessarily do them all if it is a kind of waste of time for you but please do "different-stylish" questions and explain so that I can understand on how to do them.3x^2 - 7x = 0x^2 - 14x + 49 = 09x^2 - 5x^2 = -2y(2y+9) =... 顯示更多 Please list out the steps and explain why clearly. Thanks! You don't necessarily do them all if it is a kind of waste of time for you but please do "different-stylish" questions and explain so that I can understand on how to do them. 3x^2 - 7x = 0 x^2 - 14x + 49 = 0 9x^2 - 5x^2 = -2 y(2y+9) = -9 2x^2 - 3x = x^2 + 18 (3x+2)(x-1) = 7-7x 3x^2 +33 = 2x^2 + 14x 5x^2 + 2x + 3 = 4x^2 + 6x -1 15x^2 + x = 2 (x-7)^2 = 166 4x^2 - 9 = 0 I know the answers but just need to understand why and in some parts of the equations I need to know how. Just like the second question, third and some of the others. *And that's why I request you to list out the steps. 更新: for this Q4) y(2y+9) = -9 2y^2 + 9y + 9 = 0 ( 2y+3 )( y+3 ) = 0 y = -3 or y = -3/2 can you express it more? Thanks... 更新 2: from 2y^2 + 9y + 9 = 0 why it will become (2y+3) (y+3)
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其他解答:
Quadratic Equation (Urgent)
發問:
Please list out the steps and explain why clearly. Thanks!You don't necessarily do them all if it is a kind of waste of time for you but please do "different-stylish" questions and explain so that I can understand on how to do them.3x^2 - 7x = 0x^2 - 14x + 49 = 09x^2 - 5x^2 = -2y(2y+9) =... 顯示更多 Please list out the steps and explain why clearly. Thanks! You don't necessarily do them all if it is a kind of waste of time for you but please do "different-stylish" questions and explain so that I can understand on how to do them. 3x^2 - 7x = 0 x^2 - 14x + 49 = 0 9x^2 - 5x^2 = -2 y(2y+9) = -9 2x^2 - 3x = x^2 + 18 (3x+2)(x-1) = 7-7x 3x^2 +33 = 2x^2 + 14x 5x^2 + 2x + 3 = 4x^2 + 6x -1 15x^2 + x = 2 (x-7)^2 = 166 4x^2 - 9 = 0 I know the answers but just need to understand why and in some parts of the equations I need to know how. Just like the second question, third and some of the others. *And that's why I request you to list out the steps. 更新: for this Q4) y(2y+9) = -9 2y^2 + 9y + 9 = 0 ( 2y+3 )( y+3 ) = 0 y = -3 or y = -3/2 can you express it more? Thanks... 更新 2: from 2y^2 + 9y + 9 = 0 why it will become (2y+3) (y+3)
最佳解答:
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Q1) 3x^2 - 7x = 0 x(3x-7) = 0 x = 0 or 3x-7 = 0 x = 0 or x = 7/3 Q2) x^2 - 14x + 49 = 0 x^2 - 2(7)(x) + (7)^2 = 0 (x-7)^2 = 0 x = 7 Q3) 9x^2 - 5x^2 = -2 4x^2 = -2 (2x)^2 = -2............... complex number~! recall: √(-1) = i 2x = √(-2) 2x = i√2 x = i√2 / 2 Q4) y(2y+9) = -9 2y^2 + 9y + 9 = 0 ( 2y+3 )( y+3 ) = 0 y = -3 or y = -3/2 Q5) 2x^2 - 3x = x^2 + 18 x^2 - 3x - 18 = 0 ( x-6 )( x+3 ) = 0 x = 6 or x = -3 Q6) (3x+2)(x-1) = 7-7x 3x^2 + 2x - 3x - 2 = 7 - 7x 3x^2 + 6x - 9 = 0 x^2 + 2x - 3 = 0 ( x+3 )( x-1 ) = 0 x = -3 or x = 1 Q7) 3x^2 +33 = 2x^2 + 14x x^2 - 14x + 33 = 0 ( x-11 )( x-3 ) = 0 x = 3 or x = 11 Q8) 5x^2 + 2x + 3 = 4x^2 + 6x -1 x^2 - 4x + 4 = 0 x^2 - 2(2)(x) + (2)^2 = 0 ( x-2 )^2 = 0 x = 2 Q9) 15x^2 + x = 2 15x^2 + x - 2 = 0 ( 3x-1 )( 5x+2 ) = 0 x = 1/3 or x = -2/5 Q10) (x-7)^2 = 166 x^2 - 14x + 49 = 166 x^2 - 14x - 117 = 0 using the general quadratic formula: x = ( -b√(b^2-4ac) )/2a x = ( +14√[ (-14)^2 - 4(1)(-117) ] )/2 x = ( 14√[ 196+468 ] )/2 x = ( 14 √664 ) /2 x = ( 14 2√166) / 2 x = 2( 7 √166 ) / 2 x = ( 7 √166 ) x = 7+√166 or x = 7-√166 Q11) 4x^2 - 9 = 0 4x^2 = 9 (2x)^2 = 9 2x = 3 x = 3/2 x = 3/2 or x = -3/2 2009-03-02 16:52:00 補充： For Q10)... the +- symbol cannot show in the above. So, I use +- to represent "+" and "-"... (x-7)^2 = 166 x^2 - 14x + 49 = 166 x^2 - 14x - 117 = 0 using the general quadratic formula: x = ( -b +- √(b^2-4ac) )/2a x = ( +14 +-√[ (-14)^2 - 4(1)(-117) ] )/2 x = ( 14 +-√[ 196+468 ] )/2 x = ( 14 +-√664 ) /2 2009-03-02 16:52:06 補充： x = ( 14 +- 2√166) / 2 x = 2( 7 +-√166 ) / 2 x = ( 7 +-√166 ) x = 7+√166 or x = 7-√166 2009-03-02 16:53:02 補充： For Q11)... the +- symbol cannot show in the above. ( same as Q10 ) 4x^2 - 9 = 0 4x^2 = 9 (2x)^2 = 9 2x = +-3 x = +-3/2 x = 3/2 or x = -3/2 2009-03-02 19:05:01 補充： OK... for Q4) 2y^2 + 9y + 9 = 0 you need to separate 2y^2 to 2y and y.... so, in your brain, you will have: ( 2y... )( y... ) = 0 then, you need to separate 9 to 3x3, or 1x9.... right? so again in your brain, you will have: ( 2y..3 )( y...3 ) = 0 or ( 2y...1 )( y...9 ) = 0 2009-03-02 19:06:41 補充： obviously, ( 2y..1 )( y..9 ) =0 is WRONG~! as the middle term is 9y. so, for the ( 2y..3 )( y..3 ) =0, now you need to think of the maths sign. either +, + or -, - or +, - or -, + 2009-03-02 19:07:00 補充： again, because of +9 for the equation, maths sign is either +, + or -, - so, (2y + 3)(y + 3) =0 is the correct answer.. .......Anyway, do you understand what I try to explain??其他解答:
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