HOW CAN I SOLVETHE EQUATION (X +2)(X - 3) = 14 BY CONSTRUCTING A TABLE OF VALUES!?

Question

using integer values of x between -6 and 6 x y? can someone answer this question in complete detail.
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6

Answer 1

Here's a screenshot of a spreadsheet of the values:

http://img4.imageshack.us/i/screenshot20100118at155.png/



X is across the top, and Y is down the side. A particular cell is equal to the value (x + 2)(x - 3).



So if x = 4, and y = -3, then that cell would contain the value (4 + 2)(-3 - 3) = (6)(-6) = -36



The solutions are for when the a cell value equals 14 (highlighted in orange in my picture), which is at:



(-4, -4) and (5, 5)



If you are still stuck then feel free to message me.

Answer 2

Question Number 1 :

For this function y(x)=( x + 2 ) * ( x - 3 ) , answer the following questions :

A. Looks for some points in the curve and plot the curve !



Answer Number 1 :

First, we must turn this equation ( x + 2 ) * ( x - 3 ) = 0 into a*x^2+b*x+c=0 form.

( x + 2 ) * ( x - 3 ) = 0 , expand the left hand side.

<=> x * ( x - 3 ) + 2 * ( x - 3 ) = 0

<=> x^2 - x - 6 = 0 , move 0 from the right hand side to the left hand side.

<=> x^2 - x - 6 - 0 = 0

<=> x^2 - x - 6 = 0

The equation x^2 - x - 6 = 0 is already in a*x^2+b*x+c=0 form.

As the value is already arranged in a*x^2+b*x+c=0 form, we get the value of a = 1, b = -1, c = -6.



1A. Looks for some points in the curve and plot the curve !

y(x) = x^2 - x - 6

y(-6) = (-6)^2 - (-6) - 6 = 36 -> ( (-6) , y ) = ( -6 , 36 )

y(-5) = (-5)^2 - (-5) - 6 = 24 -> ( (-5) , y ) = ( -5 , 24 )

y(-4) = (-4)^2 - (-4) - 6 = 14 -> ( (-4) , y ) = ( -4 , 14 ) <- this is the answer x=-4

y(-3) = (-3)^2 - (-3) - 6 = 6 -> ( (-3) , y ) = ( -3 , 6 )

y(-2) = (-2)^2 - (-2) - 6 = 0 -> ( (-2) , y ) = ( -2 , 0 )

y(-1) = (-1)^2 - (-1) - 6 = -4 -> ( (-1) , y ) = ( -1 , -4 )

y(0) = (0)^2 - (0) - 6 = -6 -> ( (0) , y ) = ( 0 , -6 )

y(1) = (1)^2 - (1) - 6 = -6 -> ( (1) , y ) = ( 1 , -6 )

y(2) = (2)^2 - (2) - 6 = -4 -> ( (2) , y ) = ( 2 , -4 )

y(3) = (3)^2 - (3) - 6 = 0 -> ( (3) , y ) = ( 3 , 0 )

y(4) = (4)^2 - (4) - 6 = 6 -> ( (4) , y ) = ( 4 , 6 )

y(5) = (5)^2 - (5) - 6 = 14 -> ( (5) , y ) = ( 5 , 14 ) <- this is the answer x=5

y(6) = (6)^2 - (6) - 6 = 24 -> ( (6) , y ) = ( 6 , 24 )



The answers are : x=-4 and x=5

Answer 3

You rewrite it as a quadratic and solve it that way.

(x + 2)(x - 3) = 14 is the same as x^2 - x -20 = 0

which factorises into (x + 4)(x - 5) = 0

giving as solutions x = -4 and x = +5



You might just spot that 7 times 2 = 14 in the original equation, in which case you can then

say x = 5 is one solution. Essentially it is a quadratic equation, nothing else.

Your table of values helps a bit, but won't solve the equation.

Answer 4

Instead its easier to see that 14 = 7(2) and x = 5 also 14 = (-2)(-7) so x = -4.

See if this is useful to you.

Answer 5

From the training given, i might draw out a three column table. positioned integer values i.e. -6, -5... 6 in the x column. fill in the fee of (x+2)(x-3) given x, then mark the x which will provide you 14. i don't think of you will desire to re-write this variety of the given equation.

Answer 6

36

24

14

6

0

-4

-6

-6

-4

0

6

14

24



The numbers that work are 5 and -4.