Solve By Factoring Help Need

**Ashley911** · May 16th 2010, 08:15 AM

height of ball thrown = h
t = time in seconds

h = -4.9t^2 + 38t + 1.75

what is the time of the ball when h = 50

I need to be able to solve it by factoring? i can do it if the numbers 4.9 and 38 have a common denominator but can not do when they are decimals?
any help appreciated

**e^(i*pi)** · May 16th 2010, 08:48 AM

Originally Posted by Ashley911

height of ball thrown = h
t = time in seconds

h = -4.9t^2 + 38t + 1.75

what is the height of the ball after 3 seconds

I need to be able to solve it by factoring? i can do it if the numbers 4.9 and 38 have a common denominator but can do when they are decimals?
any help appreciated

$h(t) = -4.9t^2 + 38t + 1.75$

To find the height at 3 seconds solve $h(3)$ by putting in 3 wherever you find t in the original equation. No factoring required

$h(3) = -4.9(3^2) + 38(3) + 1.75$

**Ashley911** · May 16th 2010, 08:53 AM

I have to solve it by factoring its a homework question.

**obesechicken13** · May 16th 2010, 09:24 AM

I suppose you could plug in 50 for h, then subtract 50 from both sides

should look something like
-4.9t^2 + 38t - 48.25 = 0

Then just use the quadratic formula to find the two roots of this equation. Using those two roots, you can find the factors of the polynomial. One of those roots will be your time, (the positive one).

The quadratic formula will find your factors, however this is a bit redundant, since to find the factors you will need to find the roots, and one of the roots is your answer.

Otherwise you could use the decomposition method shown here: How to Factor Second Degree Polynomials (Quadratic Equations) - wikiHow

**Ashley911** · May 16th 2010, 10:07 AM

Thanks but am still stuck here is my issue

I need to find the two numbers that when multiplied = 4.9*48.25
and
when added together are = 48.25

I keep getting decimals and believe this should not be the case.

**e^(i*pi)** · May 16th 2010, 10:19 AM

Originally Posted by Ashley911

Thanks but am still stuck here is my issue

I need to find the two numbers that when multiplied = 4.9*48.25
and
when added together are = 48.25

I keep getting decimals and believe this should not be the case.

You won't get neat factors because $b^2-4ac$ is not a perfect square

**rtblue** · May 16th 2010, 11:00 AM

$h = -4.9t^2 + 38t + 1.75$

Use the quadratic formula

$\frac{\-b\pm\sqrt{b^2-4ac}}{2a}$

plug in the values to get

$\frac{-38\pm\sqrt{38^2-4(-4.9)(1.75)}}{2(-4.90)}$

Now solve for both values: + and -

Once you have your roots then, then go ahead and solve.

**e^(i*pi)** · May 16th 2010, 11:14 AM

The OP could also complete the square - it's a kind of factoring

$50 = -4.9t^2 + 38t + 1.75$

$4.9t^2-38t = -48.25$

$t^2 - \frac{38}{4.9} = -\frac{48.25}{4.9}$

$\left(t- \frac{38}{9.8}\right)^2 - \left(\frac{38}{4.9}\right)^2 = -\frac{48.25}{4.9}$

$\left(t- \frac{38}{9.8}\right)^2 = \left(\frac{38}{4.9}\right)^2 -\frac{48.25}{4.9}$

$t - \frac{38}{9.8} = \pm \sqrt{ \left(\frac{38}{4.9} \right)^2 -\frac{48.25}{4.9}}$

$t = \frac{38}{9.8} \pm \sqrt{ \left(\frac{38}{4.9} \right)^2 -\frac{48.25}{4.9}}$

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