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Posted

Thats coz this is slow.. anyways you only said think.. I knew it was right :P

Posted

x2-3x+2=0

[x^2-2X(3/2)X(x)]+2=0

Notice that the first two terms have been brought under the structure a^2 -2ab. So you need b^2 to complete the square and here b^2-(3/2)^2 = 9/4

Therefore

[x^2-2X(3/2)X(x)+(9/4-9/4)]+2=0

[x^2-2X(3/2)X(x)+(9/4]+(-9/4+2)=0

(x-3/2)^2 = 9/4-2 = 1/4

Taking the sqroot

(x-3/2)= +1/2 or -1/2

(x-3/2)= +1/2 gives x = 1/2+3/2 = 2

(x-3/2)= -1/2 gives x = -1/2+3/2 = 1

Answer: x = 2 and x=1

solve by using the quadratic equasion

x2-3x+2=0

x2+[(-2x)+(-x)]+2=0

[splitting the middle term into two parts

whose multiplication is the product of the square term and the constant term.

-3x = (-2x) + (-x) so that (-2x)X(-x) = +(2x^2) = (x^2)X(2)]

x2-2x-x+2=0

(x2-2x)-x+2=0

x(x-2)-(x-2)=0

xp-p=0 where p = (x-2)

p(x-1) = 0

(x-2)(x-1) = 0

(x-2) = 0 gives x = 2 and (x-1) = 0 gives x = 1

Posted

Well done

x = -1 or -2 both work

How bout this

Me + bed = x

x = sleep :)

night night all

  • 2 weeks later...

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