# Thread: Easy (and short) problem

1. ## Easy (and short) problem

The amount of bending, B mm, of a particular wooden beam under a load is given by B = 0.2m^2 + 0.5m +2.5, where m kg is the mass (or load) on the end of the beam. What mass will produce a bend of 8.8 mm?

I tried to substitute 8.8 for B and then subtracted it from both sides of the equation. After this I tried to use the quadratic formula but was unsuccessful.

Please provide complete working out.

2. Originally Posted by Joker37
The amount of bending, B mm, of a particular wooden beam under a load is given by B = 0.2m^2 + 0.5m +2.5, where m kg is the mass (or load) on the end of the beam. What mass will produce a bend of 8.8 mm?

I tried to substitute 8.8 for B and then subtracted it from both sides of the equation. After this I tried to use the quadratic formula but was unsuccessful.

Please provide complete working out.
Post your working so we can show you where you went wrong, since you have the correct idea on how to do this problem.

3. The way I'd solve it is:

$88=2m^2+5m+25$

$2m^2+5m-63=0$

$(2m-9)(x+7)=0$

$\therefore m=4.5 kg$

As $m$ can't be negative

4. Originally Posted by Chris L T521
Post your working so we can show you where you went wrong, since you have the correct idea on how to do this problem.
B = 0.2m^2 + 0.5m + 2.5
8.8 = 0.2m^2 + 0.5m + 2.5
0.2m^2 + 0.5m - 6.3 = 0

Using Quadratic formula = confusing result
(sorry I can't use LaTex)

5. Worked for me when I used the quadratic formula:

$\frac{-0.5+\sqrt{0.5^2-4(0.2)(-6.3)}}{2(0.2)}=4.5$