jeffry is wrong because it only works for 4 let's say jef was right we would get (4)(4)=4, then 16=4 which is false, go back to school, jeff we will solve to find out if george is right later (he's wrong because he set it 0 when it is equal to 4) anwyay, the way to do it is expand, make one side 0 and then factor, then set each factor to 0 so expanding we get 2x²+5x-3=4 minus 4 both sides 2x²+5x-7=0 ok, we need to use the ac method or something for ax²+bx+c=0 where a is not eual to 0, multiply a and c and call the result r, find what 2 numbers multiply to get r and add to get b in the equation 2x²+5x-7=0 2 times -7=-14 what 2 numbers multiply to get 5 and add to get -14 -2 and 7 split middle number up 2x²-2x+7x-7=0 group (2x²-2x)+(7x-7)=0 factor 2x(x-1)+7(x-1)=0 undistribute (2x+7)(x-1)=0 set each to 0 2x+7=0 2x=-7 x=-7/2 x-1=0 x=1 x=-7/2 and 1 that's how it works to check x=-7/2 (2(-7/2)-1)((-7/2)+3)=4 (-7-1)((-7/2)+(6/2))=4 (-8)(-1/2)=4 8/2=4 4=4 true x=1 (2(1)-1)(1+3)=4 (2-1)(4)=4 (1)(4)=4 4=4 true x=-7/2 and 1 george and jeff are wrong

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