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secondary 3 | E Maths
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I've managed to form the right quadratic equation. I can get the correct answers using the multiplication frame, but I can't get the answer m by completing the square. What am I doing wrong here?
Date Posted: 3 years ago
Views: 263
What was the original question?
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My advise is only do completing the square when the coefficient of x^2 is +1.
Suggested Solutions (Step-By-Step)
Do drop me a sms if you need further clarifications. Hope this helps!
Suggested Solutions (Step-By-Step)
Do drop me a sms if you need further clarifications. Hope this helps!
Date Posted:
3 years ago
Hi! Does the coefficient have to be +1 for x², or can it be any other positive number? Also if we are left with a negative coefficient, can I factor out the negative sign and still use completing the square method? Thanks!
Hi yes I am assuming the (x + 7) in your first line was meant to be bracketed.
Also, yes the coefficient of x² needs to be +1 before commencing completing the square. If it is a positive number, just divide throughout both LHS and RHS by that positive number to obtain +1 as the coefficient of x². This is the same for any negative coefficient of x². Just divide throughout both LHS and RHS by that negative number to obtain +1 as the coefficient of x². Then proceed to complete the square as required.
The above technique is to be used only if it is an equation with RHS equal to zero, e.g.,
2x² + 3x - 7 = 0
.
.
.
If it is just an expression, e.g.,
y = 2x² + 3x - 7
DO NOT divide throughout by 2. Instead, just factorise out the 2, i.e.,
y = 2 [ x² + (3x)/2 - 7/2 ]
Then proceed to complete the square for the things inside the square brackets as usual.
Hope this helps! Do let me know if you need further clarifications! : )
Also, yes the coefficient of x² needs to be +1 before commencing completing the square. If it is a positive number, just divide throughout both LHS and RHS by that positive number to obtain +1 as the coefficient of x². This is the same for any negative coefficient of x². Just divide throughout both LHS and RHS by that negative number to obtain +1 as the coefficient of x². Then proceed to complete the square as required.
The above technique is to be used only if it is an equation with RHS equal to zero, e.g.,
2x² + 3x - 7 = 0
.
.
.
If it is just an expression, e.g.,
y = 2x² + 3x - 7
DO NOT divide throughout by 2. Instead, just factorise out the 2, i.e.,
y = 2 [ x² + (3x)/2 - 7/2 ]
Then proceed to complete the square for the things inside the square brackets as usual.
Hope this helps! Do let me know if you need further clarifications! : )
Thanks!
Most welcome! Glad to help!
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Learn to spot mistakes and it will save a lot of unnecessary time.
When you find answer a bit odd, check first!
When you find answer a bit odd, check first!
Date Posted:
3 years ago
Ok thanks
The “x + 7” was probably meant to be bracketed, such that his working becomes
(x + 7) (x - 2) + ...
(x + 7) (x - 2) + ...
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