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y2 - 5y = -4

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y2 - 5y = -4

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Asked 6/20/2014 8:33:55 PM

Updated 6/24/2014 7:38:40 PM

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Flagged by yeswey [6/24/2014 7:38:07 PM]

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User: y2 - 5y = -4

Weegy: solve y2 - 5y = 3 ? y=5.541 , [ y=-0.5414 Solution for y2-5y=3 //- - - - - - // Solutions : Two solutions were found //- - - - - - y = -0.541 y = 5.541 Tiger made the following changes to your input: (1): "y2" was replaced by "y^2" Note - If you insist on having y*2, please input explicitly "y*2" "2y" however,retains the "2*y" meaning These changes do not affect the solution //- - - - // Drill : y^2-5y=3 //- - - - Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation : y^2-5*y-(3)=0 Solve : ((y2) - 5y) - 3 = 0 //- - - - - - // Step # 01: Simplify y2-5y - 3 //- - - - - - # 01.01 Factoring y2-5y-3 Method : Factoring by splitting the middle term The first term is y2 ,its coefficient is 1 The middle term is -5y ,its coefficient is -5 The last term, also called 'the constant', is -3 Step-1 : Multiply the coefficient of the first term by the constant 1 ? -3 = -3 Step-2 : Find two factors of -3 whose sum equals the coefficient of the middle term, which is -5 -3 + 1 = -2 -1 + 3 = 2 Observation: No such two factors can be found !! Conclusion : Trinomial can not be factored Equation at the end of step 01 : y2 - 5y - 3 = 0 //- - - - - - // Step # 02: Solve y2-5y-3 = 0 //- - - - - - # 02.01 Solving y2-5y-3 = 0 by Completing The Square Add 3 to both side of the equation : y2-5y = 3 Now the clever bit: Take the coefficient of x, which is 5, divide by two, giving (5/2), and finally square it giving (25/4). Add (25/4) to both sides of the equation : ]
jeffreymcmillan|Points 680|

User: solve the quadratic equation

Weegy: SOLVING QUADRATIC EQUATIONS BY THE NEW "TRANSFORMING METHOD". [ It uses 3 features in its solving process: - The Rule of Signs for Real Roots of a quadratic equation to look for a better solving approach. - The Diagonal Sum Method to solve simplified quadratic equations type x^2 + bx + c = 0, when a = 1. This method can immediately obtain the 2 real roots of the equation. - The transformation of a quadratic equation in standard form ax^2 + bx + c = 0 into the simplified form, with a = 1, to make the solving process much easier. ]
elionor|Points 45|

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Asked 6/20/2014 8:33:55 PM

Updated 6/24/2014 7:38:40 PM

1 Answer/Comment

This conversation has been flagged as incorrect.

Flagged by yeswey [6/24/2014 7:38:07 PM]

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yeswey

y^2 - 5y = -4
y^2 - 5y + 4 = 0
(y - 1)(y - 4) = 0
y = 1 or y = 4

Added 6/24/2014 7:38:40 PM

This answer has been confirmed as correct and helpful.

Confirmed by jeifunk [6/24/2014 10:58:46 PM]

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Updated 6/20/2014 6:34:24 PM

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jerry06

The slope between points (2, 3) and (-1, 0) is 1. m = (0-3)/(-1-2); m = -3/-3 = 1

Added 6/20/2014 6:34:21 PM

This answer has been confirmed as correct and helpful.

Confirmed by andrewpallarca [6/21/2014 1:49:26 AM]

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Updated 6/20/2014 8:35:27 PM

1 Answer/Comment

jerry06

x^2 - x - 90 = 0 = (x - 10)(x + 9) = 0

Added 6/20/2014 8:35:25 PM

This answer has been confirmed as correct and helpful.

Confirmed by jeifunk [6/20/2014 8:37:05 PM]

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Question

Updated 6/22/2014 1:52:56 PM

1 Answer/Comment

andrewpallarca

m^2+ 2m - 24 = (m + 6)(m - 4);

m + 6 = 0;
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Added 6/22/2014 1:52:55 PM

This answer has been confirmed as correct and helpful.

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Question

Updated 6/20/2014 8:52:24 PM

2 Answers/Comments