# Quadratics And Algebra Products Help!

• Aug 31st 2008, 05:30 PM
satishgaire
Hi, I am new to this forum. I am going to be very active on this forum as i will need helps sometimes and give helps to others when i know.

Please do explain how you did it.
Its a quadratic question. I know how to solve a regular quadratics problem but this one looks little different.

y^2-24y+144=0

Please explain how so that i can do other problems.

and few simple products for algebra

(t+7)^2

(c-2d)^2

All helps are appreciated. Thank you.
• Aug 31st 2008, 05:46 PM
Jhevon
Quote:

Originally Posted by satishgaire
Hi, I am new to this forum. I am going to be very active on this forum as i will need helps sometimes and give helps to others when i know.

Please do explain how you did it.
Its a quadratic question. I know how to solve a regular quadratics problem but this one looks little different.

y^2-24y+144=0

Please explain how so that i can do other problems.

we can factorize this. think of two numbers that we can multiply to get +144 and add to get -24. the two numbers are -12 and -12

thus we can set up our brackets and place these numbers in the proper places:

$(y - 12)(y - 12) = 0$

$\Rightarrow (y - 12)^2 = 0$

$\Rightarrow y - 12 = 0$ ...........i square rooted both sides

$\Rightarrow y = 12$

alternatively, if you can't factorize, use the quadratic formula.

the solutions to $ax^2 + bx + c = 0$ are given by the formula:

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

in this problem, a = 1, b = -24 and c = 144

Quote:

and few simple products for algebra

(t+7)^2

(c-2d)^2

you can use the formula: $(a + b)^2 = a^2 + 2ab + b^2$

example: $(t + 7)^2 = t^2 + 2 \cdot t \cdot 7 + 7^2 = t^2 + 14t + 49$

the longer way: write it out as (t + 7)(t + 7). now multiply everything in the second set of brackets by everything in the first set.

(t + 7)(t + 7) = t*t + t*7 + 7*t + 7*7 = $t^2 + 14t + 49$