• Oct 15th 2007, 10:55 AM
IGCSE student
Hi there,
There are some questions in my homework which i totally dont understand.
This homework is due in wednesday 8:30am.

here they are

Solve by forming a quadratic equation

1) The length of a rectangle exceeds the width by 2cm. If the diagonal is 10cm long, find the width of the rectangle. ( use Pythagoras' Theorem)

2) The perimeter of a rectangle is 68cm. I f the diagonal is 26cm, find the dimensions of the rectangle.

3) A man walks a certain distance due North and then the same distance plus a futher 7km due east. If the final distance for the starting point is 17km, find the distance he walks from North to East.

4)A boy buys x eggs at (x-8) cents on each and (x-2) rashers of bacon at (x-3) cent each. If the total bill was $1.75, how many eggs does he buy? 5) A number exceeds four times its reciprocal by 3. Find the number Thanks • Oct 15th 2007, 11:19 AM CaptainBlack Quote: Originally Posted by IGCSE student Hi there, There are some questions in my homework which i totally dont understand. This homework is due in wednesday 8:30am. here they are Solve by forming a quadratic equation 1) The length of a rectangle exceeds the width by 2cm. If the diagonal is 10cm long, find the width of the rectangle. ( use Pythagoras' Theorem) Always start by assigning the variables in the problem names. Let the length be x, and the width be y. Then we are told that x=y+2, and the diagonal is 10cm, so by Pythagoras' theorem: 100 = x^2+y^2 = (y+2)^2+y^2, so: 2y^2 + 4y - 96 = 0 which is for you to now solve. RonL • Oct 15th 2007, 11:33 AM IGCSE student what is the best method i could use to solve the equation • Oct 15th 2007, 11:38 AM topsquark Quote: Originally Posted by IGCSE student what is the best method i could use to solve the equation Factoring, if it works, is probably the best way. If it doesn't factor, I recommend completing the square as a good exercise, but if you need the answer fast, you can use the quadratic equation. -Dan • Oct 15th 2007, 11:44 AM IGCSE student ac have any of u heard of the ac method? its a new method that was introduced last year, and its being used in the IGCSE coarse. • Oct 15th 2007, 12:44 PM topsquark Quote: Originally Posted by IGCSE student have any of u heard of the ac method? its a new method that was introduced last year, and its being used in the IGCSE coarse. It is a method used to factor quadratics. (And it isn't new. It's pretty old, actually.) Consider the following quadratic:$\displaystyle 2x^2 - x - 6$Multiply the coefficient of the$\displaystyle x^2$term by the constant term. In this case,$\displaystyle 2 \cdot -6 = -12$. Now write all the pairs of factors of this number: 1, -12 2, -6 3, -4 4, -3 6, -2 12, -1 Now you look for a pair of those factors that add up to the coefficient of the$\displaystyle x$term, in this case -1: 3 + -4 = -1 (If you cannot find such a pair, the quadratic does not factor over the rationals and you have to use another method.) So rewrite the linear term of the quadratic using this pair of numbers:$\displaystyle 2x^2 - x - 6 = 2x^2 + (3x - 4x) - 6$Now regroup it like this:$\displaystyle (2x^2 + 3x) + (-4x - 6)$and finally, factor it by grouping:$\displaystyle (2x^2 + 3x) + (-4x - 6) = x(2x + 3) - 2(2x + 3)\displaystyle = (x - 2)(2x + 3)$Now you apply this method to solve$\displaystyle 2y^2 + 4y - 96 = 0\$. (Hint: Factor out the 2 common to all terms first.)

-Dan
• Oct 15th 2007, 04:16 PM
IGCSE student
Thanks for explaining the AC method to thoes whom dnt know, personally i understand the method completely.