question (1) - solve 4y2 + 8y + 3 = 0
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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 |
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[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 |
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[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 |
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( 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 |
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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 |
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(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(+/-) ). |
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(2y+3)(2y+1)= 0 |
[9] |
Now set each factor or group in equation obtained in step[8] equal to zero |
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2y+3 = 0 and 2y+1 = 0 |
[10] |
Now solve both equations obtained in step[9] |
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2y+3 = 0
2y = 0 - 3
2y = - 3
y = -3/2 |
and |
2y+1 = 0
2y = 0 - 1
2y = -1
y = -1/2
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Hence the solutions to given equation are |
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