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Solving Quadratic Equations
By Completing the square / Exercises

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question (2) - solve 5y2 + 12y + 4 = 0
Step Explanation pre-result result after applying step
[0] writing given equation 5y2 + 12y + 4 = 0
[1] Substract constant term '4' from both sides of equation
and simplifying
5y2 + 12y + 4 - 4 = 0 - 4
5y2 + 12y = -4
[2] Now divide the equation obtained in previous step throughout by the coefficient of y2 or the number in front of y2 ,which is 5 in this case
and simplifying
i.e (5y2/5) + 12y/5 = -4/5
y2 + (12/5)y = -4/5
[3] Now find half of '12/5', the coefficient of y or find half of the number in front of y of the last equation obtained in step(2)
, square the result and add it to both sides of equation
(1/2)x(12/5)
= 6/5
62/52
y2 + (12/5)y + 62/52= -4/5 + (6/5)2
[4] Since (6/5)+(6/5)= 12/5
and (6/5)X(6/5)= 62/52
we factor the left hand side of last equation obtained in step(3)
to factor all we do is take the variable which is y ,in this case, and add it to the the constant '6/5' that is squared on the left hand side of previous equation, and square the result
blank [ y + (6/5)]2= (-4/5)+(6/5)2
[5] Simplify right hand side of equation since [ y + (6/5)]2= (-4/5)+(6/5)2
[ y + (6/5)]2 = (-4/5)+(36/25)
[ y + (6/5)]2 = (-20 + 36)/25
[ y + (6/5)]2 = 16/25
[6] Now solve the last equation obtained in step(5) for y,
by taking the square root on both sides of equation now since the square root of any number R2 is that number which when multiplied by itself gives R2
note that R x R = R2 , and
-R x -R = R2
so, the square root of R2 is R and -R ,thus the square root of a number includes the negative root also ,for example the square roots of 25 are 5 and -5 ,since 5x5=25 and -5x-5=25.
Since the squareroot of a number squared is the number itself ,the squareroot [ y + (6/5)]2 is y + (6/5) y +(6/5)= (+/-) Ö(16/25)
[7] Since Ö(16/25) = (+/-)4/5 ,
simplifying the right hand side of the previous equation gives:
blank y+(6/5) = (+/-)4/5
[8] (+/-) implies that blank y+(6/5) = +(4/5) and y+(6/5)= -(4/5)
[9] Now solve both equations obtained in step[8] blank
y+(6/5) = 4/5
y = (4/5)-(6/5)
y = - 2/5
and y+(6/5) = -4/5
y = (-4/5)-(6/5)
y = -10/5
y = -2
Hence the solutions to given equation are blank
y = -2/5 and y = -2

© 2004
Prakash Sukhu