find the roots of the following quadratic equations by factorisation non of the dinometer is zero . 1 upon x-2 + 2 upon x -1 =6 upon x
Answers
Answer:
It’s Professor Dave, let’s complete the
square.
We just learned how to solve polynomials by
factoring, but not all polynomials can be
factored, so we definitely need a few other
techniques.
One technique that will always work is called
completing the square, so let’s see how
that works now.
First consider the following polynomial, X
squared plus two X plus one.
This can be factored, but furthermore, it
can be factored into two identical binomials.
The factors of one are one and one, or negative
one and negative one, and the positive ones
do indeed add up to two, so we can factor
this into X plus one times X plus one.
Anything times itself is by definition that
thing squared, so we can also express this
as X plus one quantity squared.
If this were equal to zero, the solution is
very simple, it’s just negative one, because
that’s the value for X that will make this
equal to zero.
So the original polynomial must be a perfect
square, since it can be generated by squaring
this binomial, just like four is a perfect
square because it’s two squared.
This kind of thinking comes in handy for certain
polynomials, because while they might not
factor, we can deliberately add some term
to the expression that will manufacture a
polynomial that is a perfect square, thus
allowing us to express it as the square of
a binomial, and making it easy to solve.
This will make more sense with an example.
Let’s say we have X squared plus two X minus
six equals zero.
If we find the factors of negative six, which
are combinations of one and six or two and
three, we find that there is no way to add
any pair of factors together to get positive
two, so we can’t factor this polynomial.
But let’s do a little trick.
Let’s add six to both sides, bringing the
six to the other side.
Earlier we saw that X squared plus two X plus
one is a perfect square, so what if we now
add one so that we suddenly do have a perfect
square polynomial, in essence completing the
square?
Well whatever we do, we have to do it to both
sides, so we end up with X squared plus two
X plus one equals seven.
The reason this is useful, is that we can
now express the left side as X plus one quantity
squared.
From here, it’s not very hard to solve.
We have an exponent of two, so we take the
square root of both sides.
That gives us X plus one equals plus or minus
root seven.
Remember that when taking the square root
of a number, we get two answers, since both
of these square to give seven
Answer:
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