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secondary 4 | E Maths
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Pls help me with q11(a). Thanks!
Date Posted: 4 years ago
Views: 222
2(a^2-5)=8
2a^2-10=8
2a^2=18
a^2=9
Due to a is integer and a>1
So a=3
2a^2-10=8
2a^2=18
a^2=9
Due to a is integer and a>1
So a=3
Your approach is incomplete, but answer is ok.
Because (a^2 - 5)^2 is 16, the a^2 - 5 can take on 4 or -4.
The first leads to a^2 = 9.
The second leads to a^2 = 1.
a can take on values -3, -1, 1 or 3, but because a > 1, a can only be 3.
Because (a^2 - 5)^2 is 16, the a^2 - 5 can take on 4 or -4.
The first leads to a^2 = 9.
The second leads to a^2 = 1.
a can take on values -3, -1, 1 or 3, but because a > 1, a can only be 3.
Why is (a^2-5)^2 16?
Let the function be called y, for simplicity.
y = x2 - 8x + (a^2 - 5)^2.
Let’s ignore the fact that (a^2 - 5)^2 exists for the time being.
y = x^2 - 8x
There is the completing the square method. There are other similar, but related, methods too.
The nice thing about a completed square form is that
(a + b)^2
= a^2 + ab + ab + b^2
= a^2 + 2ab + b^2
In our case, it’s y = x^2 - 8x. We still ignore the third term for now.
We can rewrite this as
y = x^2 - 4x - 4x
Doesn’t this resemble a^2 + ab + ab + b^2? But it’s missing the last term, but at least they look similar. If you have observed carefully, x is like the “a” and -4 is like the “b”.
The missing term is “b^2” which translates to (-4)^2, or 16.
Now look at the original expression
y = x^2 - 8x + (a^2 - 5)^2
y = x^2 - 4x - 4x + (a^2 - 5)^2
Previously we mentioned that the missing term is 16. This must be the value of (a^2 - 5)^2.
Check that y = x^2 - 8x + 16 can be factorised into (x - 4) (x - 4) which is simply a perfect square (x - 4)^2.
Hence,
(a^2 - 5)^2 = 16
(This can be seen from the cross factorisation table too, but in a slightly different way from the usual factorisation of quadratic expressions)
We do not need the (a^2 - 5) to be 4 all the time, but we need the (a^2 - 5)^2 to be 16 all the time.
Square rooting both sides,
a^2 - 5 = 4 ### OR ### a^2 - 5 = -4
a^2 = 9 ### OR ### a^2 = 1
We will obtain four values for a, namely -3, -1, 1 and 3, but given the extra condition that a > 1, the only value of a we can accept is 3.
y = x2 - 8x + (a^2 - 5)^2.
Let’s ignore the fact that (a^2 - 5)^2 exists for the time being.
y = x^2 - 8x
There is the completing the square method. There are other similar, but related, methods too.
The nice thing about a completed square form is that
(a + b)^2
= a^2 + ab + ab + b^2
= a^2 + 2ab + b^2
In our case, it’s y = x^2 - 8x. We still ignore the third term for now.
We can rewrite this as
y = x^2 - 4x - 4x
Doesn’t this resemble a^2 + ab + ab + b^2? But it’s missing the last term, but at least they look similar. If you have observed carefully, x is like the “a” and -4 is like the “b”.
The missing term is “b^2” which translates to (-4)^2, or 16.
Now look at the original expression
y = x^2 - 8x + (a^2 - 5)^2
y = x^2 - 4x - 4x + (a^2 - 5)^2
Previously we mentioned that the missing term is 16. This must be the value of (a^2 - 5)^2.
Check that y = x^2 - 8x + 16 can be factorised into (x - 4) (x - 4) which is simply a perfect square (x - 4)^2.
Hence,
(a^2 - 5)^2 = 16
(This can be seen from the cross factorisation table too, but in a slightly different way from the usual factorisation of quadratic expressions)
We do not need the (a^2 - 5) to be 4 all the time, but we need the (a^2 - 5)^2 to be 16 all the time.
Square rooting both sides,
a^2 - 5 = 4 ### OR ### a^2 - 5 = -4
a^2 = 9 ### OR ### a^2 = 1
We will obtain four values for a, namely -3, -1, 1 and 3, but given the extra condition that a > 1, the only value of a we can accept is 3.
See 1 Answer
X² - 8X + (a² - 5)² is a perfect square, meaning that it can be written in the form of ( X - (a² - 5) )² or ( (a² - 5) - X )²
As such, there are two possible values of a.
We know frow that the expansion of (a - b)² results in a² - 2ab + b² , and the -8 in the question is the "-2ab"
So -2 (a²-5) X is equal to -8X and (a²-5) = 4 or -4
There are two values of a that satisfy this,
a = 1 and a = 3
when a = 1, (a²-5) = -4.
when a = 3, (a²-5) = 4
As such, there are two possible values of a.
We know frow that the expansion of (a - b)² results in a² - 2ab + b² , and the -8 in the question is the "-2ab"
So -2 (a²-5) X is equal to -8X and (a²-5) = 4 or -4
There are two values of a that satisfy this,
a = 1 and a = 3
when a = 1, (a²-5) = -4.
when a = 3, (a²-5) = 4
Actually four values -3, -1, 1 and 3, and there is the given condition that a > 1 from the question.
Hence, a = 3.
Hence, a = 3.
Yep! Still have to reject the value of a = 1.
Didn't bother considering a of negative values as we can see that a won't be negative from the conditions.
Didn't bother considering a of negative values as we can see that a won't be negative from the conditions.
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