# solving this equation problem

• Dec 3rd 2005, 12:31 PM
rpatel
solving this equation problem
so far i have solved the answer to part a).

but i can't seem to solve the equation for part b) though. to solve this qudractic do i have to use the qaudractic formula?

any help is appreichated thank you.

:D
• Dec 3rd 2005, 02:57 PM
Rich B.
Hi Rpatel:

Strictly speaking, the quadratic formula is nor required in this instance because the trinomial 10x^2-11x-84 is factorable. FYI, any quadratic trinomial, ax^2+bx+c, is factorable whenever its discriminant, b^2-4ac, is a perfect square. In the present case, b^2-4ac = (-11)^2 - 4(10)(-84) = 3481, which is indeed a perfect square (Any integer, k, is a perfect square if sqrt(k) is itself an integer. Because sqrt(3481)=59, and 59 is an integer, it follows that 3481 is by definition a perfect square). Nevertheless, were this my problem, I would use the q.formula just the same, as the coefficients 10 and 84 make for more factor combinations than I care to test.

Regards,

Rich B.
• Dec 3rd 2005, 04:14 PM
ticbol
Quote:

Originally Posted by rpatel
so far i have solved the answer to part a).

but i can't seem to solve the equation for part b) though. to solve this qudractic do i have to use the qaudractic formula?

any help is appreichated thank you.

:D

There is someting wrong.

Your 10x^2 -11x -84 = 0 is correct per shown figure.
And that gives x = 3.5cm or -2.4cm

x=3.5 fails at x-7.
3.5 -7 = -3.5cm ------cannot be. No negative dimensions.

x = -2.4m ------cannot be.
• Dec 3rd 2005, 10:40 PM
CaptainBlack
Quote:

Originally Posted by Rich B.
Hi Rpatel:

Strictly speaking, the quadratic formula is nor required in this instance because the trinomial 10x^2-11x-84 is factorable.

In what sense? By the Fundamental Theorem of Algebra every
quadratic has a root, and so is factorisable. It's just that these
roots may not be real, or rational.

Quote:

FYI, any quadratic trinomial, ax^2+bx+c, is factorable whenever its discriminant, b^2-4ac, is a perfect square.
Discriminant Positive - if we are interested in real factors.

For instance the discriminant of: $x^2-2$ is 8, which
is not a perfect square, but that does not mean we cannot factorise it:

$x^2\ -\ 2\ =\ (x-\sqrt2)(x+\sqrt2)$.

When the discriminant is a perfect square it does ensure that
the roots are rational.

Quote:

In the present case, b^2-4ac = (-11)^2 - 4(10)(-84) = 3481, which is indeed a perfect square (Any integer, k, is a perfect square if sqrt(k) is itself an integer. Because sqrt(3481)=59, and 59 is an integer, it follows that 3481 is by definition a perfect square). Nevertheless, were this my problem, I would use the q.formula just the same, as the coefficients 10 and 84 make for more factor combinations than I care to test.
Quote:

Regards,

Rich B.
By the time you have computed the discriminant and determined it
is a perfect square you have done almost all the work involved
in finding the roots using the quadratic formula and may as well
go ahead and find them that way, and then complete the factorisation,
or not (in this case we are interested in the roots not the factors).

RonL