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How to solve the equations:

0 votes

a) 3 x (2x + 5) = 0

b) 2x^ 2 – 16 x = 18?

asked Jun 25, 2014 in ALGEBRA 1 by anonymous

1 Answer

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a) The equation 3x ( 2x + 5) = 0

Apply zero product property .

3x = 0 and 2x + 5 = 0

x = 0 and 2x = -5

x = 0 and x  = -5/2

Solutions are image

b) The equation 2x2 - 16x = 18

2x2 - 16x - 18 = 0

  • Now factorise the  above equation.

Multiply first term 2x2 and last term -18 = -18x2

The correct pair of the terms -18x and 2x multiply to -18x2 and add to -16x.

Replace the middle term -16x with - 18x + 2x.

2x2 - 18x + 2x - 18 = 0

Group the terms into two pairs.

(2x2 - 18x) + (2x - 18) = 0

Factor out 2x from the first group  and factor out 2 from the second group.

2x(x - 9) + 2(x - 9) = 0

Factor out common term x - 9.

(x - 9) (2x + 2) = 0

x -9 = 0 and 2x + 2 = 0

x = 9 and 2x = -2

x = 9 and x = -1

Solution are x = 9 , -1.

answered Jun 25, 2014 by david Expert

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