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Solving Quadratic Equations
By Factoring / Exercises

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question (1) - solve 4y2 + 8y + 3 = 0
Step Explanation pre-result result after applying step
[0] writing given equation 4y2 + 8y + 3 = 0
[1] Multiply coefficient of y2which is '4' by the constant term '3' 4 x 3 = 12 blank
[2] Identify the coefficient of y or the number in front of y which is '8' in this case then look for two numbers whose sum is 8 and whose product is 12 6 x 2 = 12
6 +2 = 8
blank
[3] Since 6 and 2 are the two numbers we were looking for in the previous step and since y is the variable of the equation ,we multiply both 6 and 2 by y ,i.e
6 x y = 6y
2 x y = 2y
now since 6y + 2y = 8y we replace 8y with 6y + 2y
8y = 6y + 2y 4y2 + 6y + 2y + 3 = 0
[4] Now put the terms into groups of two blank ( 4y2 + 6y ) + ( 2y + 3 )= 0
[5] Factorise each group ,choosing the HCF of each group ,putting the HCF of each group outside the brackets dividing both terms in the group by the HCF of that group HCF of 4y2 and 6y is 2y
and the HCF of 2y and 3 is 1
2y[(4y2/2y) + (6y/2y)] + 1[(2y/1) + (3/1)]= 0
[6] Simplify left hand side of equation obtained in previous step blank 2y(2y + 3) + 1(2y + 3)= 0
[7] now since 2y+3 is a common factor of the terms on left hand side of equation obtained in step 6 ,factor out 2y+3 blank (2y+3)[(2y(2y + 3)/(2y+3))+ (1(2y + 3)/(2y+3))]= 0
[8] Simplify left hand side of equation obtained in previous step
Note: in the previous step you donot have to show how you factorise 2y+3 ,
you can just take (2y+3) since it appears in both terms in step [6] and multiply it to the sum of the numbers outside each brackets (paying attention to their sign(+/-) ).
blank (2y+3)(2y+1)= 0
[9] Now set each factor or group in equation obtained in step[8] equal to zero blank 2y+3 = 0 and 2y+1 = 0
[10] Now solve both equations obtained in step[9] blank
2y+3 = 0
2y = 0 - 3
2y = - 3
y = -3/2
and 2y+1 = 0
2y = 0 - 1
2y = -1
y = -1/2
Hence the solutions to given equation are blank
y = -3/2 and y = -1/2

© 2004
Prakash Sukhu