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Mathematics 13 Online
OpenStudy (anonymous):

PLEASE HELPPPPPPPPPPP =( (2x-1)^2+5(2x-1)=24

OpenStudy (anonymous):

do you have any idea what to do or how to start?

OpenStudy (anonymous):

you will be using the distributive property to start to expand the equation and than combine like terms and solve for x

OpenStudy (anonymous):

no

OpenStudy (anonymous):

it didn't work out

OpenStudy (anonymous):

first of all expand (2x-1)(2x-1) can you distribute this?

OpenStudy (anonymous):

yea but im telling u it doesn't work

OpenStudy (anonymous):

4x^2-2x-2x+1 = 4x-4x+1

OpenStudy (anonymous):

4x^2-4x+1

OpenStudy (anonymous):

correct that is the first part now distribute 5(2x-1)

OpenStudy (anonymous):

10x-5

OpenStudy (anonymous):

ok so if you add them together what do you have?

OpenStudy (anonymous):

what about the 24

OpenStudy (anonymous):

we are just working with the left side. whatever you get on the left side should = 24 so just be carrying the 24 down

OpenStudy (anonymous):

4x^2+6x-5

OpenStudy (anonymous):

-4

OpenStudy (anonymous):

correct your equation should be 4x^2+6x-4=24

OpenStudy (anonymous):

-28?

OpenStudy (anonymous):

yes 4x^2+6x-28=0 any idea whats next?

OpenStudy (anonymous):

nope

OpenStudy (anonymous):

write it in factor form

OpenStudy (anonymous):

depends how you learn it, you can use the quadratic formula but it thinks its longer or just write it in factor form, do you know how do any?

OpenStudy (anonymous):

\[ax ^{2}+bx-c=(a+b)(a-b)\] is this true?

OpenStudy (anonymous):

no i dont can u help me?

OpenStudy (anonymous):

correct, so if you do what what would you get?

OpenStudy (anonymous):

ok well you would take the 4x^2+6x-28=0 I divided the whole equation by two to reduce the large numbers which = this 2x^2+3x-14=0 than factor it to (x-2)(2x+7)=0

OpenStudy (anonymous):

whoa...

OpenStudy (anonymous):

did i confuse you?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

sorry, can you see that the whole equation is divisible by 2? i just did this to reduce the large numbers.

OpenStudy (anonymous):

okay gotcha

OpenStudy (anonymous):

ok good, and factoring it out i used the equation you showed me earlier, ax2+bx−c=(a+b)(a−b) to get the (x-2)(2x+7)=0.

OpenStudy (anonymous):

sorry don't see why it shows like that but i used the equation you showed me earlier to factor the equation 2x^2+3x-14=0

OpenStudy (anonymous):

i see nothing but little squares

OpenStudy (anonymous):

use the equation ax^2+bx-c=(a+b)(a-b) to factor the 2x^2+3x-14=0 equation.

OpenStudy (anonymous):

i got x^2=3 and x^2=4

OpenStudy (anonymous):

oops 9 not 3

OpenStudy (anonymous):

um not quite, a*a should equal your first ten which should be 2 so numbers multiply that equal 2 are what? 1 and 2 correct?

OpenStudy (anonymous):

*term

OpenStudy (anonymous):

b*b should equal your 3rd term, and the product a*b+b*a should equal your middle term.

OpenStudy (anonymous):

im confused im just gonna go to the school tutor

OpenStudy (anonymous):

it is difficult concept but if you factor it you will get (x-2)(2x+7)=0 and if you set each =0 like this: x-2=0 2x+7=0 and solve for x in each equation you will get: x=2 x=-7/2

OpenStudy (anonymous):

goodluck! :)

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