Maths Question?

Question

How do I get from x(x+1)=42 to x=6
I'm really stumped. I worked out the answer but I cant understand how to get the answer through solving the equation...

Answer 1

use the quadradic equation. Put the equation in its proper form = x^2 + x -42 = 0

the constants are
a=1
b=1
c= - 42.
Solve for the roots

remember that the quadratic equation is:
x= (-b +/- (b^2-4ac)^1/2)/2a

Answer 2

x(x+1)=42

First multiply LHS by 'x'.
x^2 + x = 42
Next subtract 42 from both sides

x^2 + x - 42 = 0

Factorise this quadratic eq'n, that is split it up in such as way that it can be put into two sets of brackets and equate to zero.
In doing this, find two numbers which multiply together to give '42' and when 'added' give '1'.
This is done by factorising :-
42 x 1 =42
21 x 2 = 42
14 x 3 = 42
7 x 6 = 42
NB taking the last multiple (7 x 6) the two numbers '6' & '7' have a difference of one '1'. So:-
(x + 7)(x - 6) = 0
NB when '+7' and '-6' are multipliied together the answer is '-42'. However '7 - 6' = +1(+x) in the eq'n.
Since any term when mulyiplied by zero = zero we can write.
x + 7 = 0 => x = -7
and/0r
x - 6 = 0 => x = 6 - your answer
So 'x' has two values '+6' and '-7'.

Hope this helps!!!!

Answer 3

if you put 6 in where the x is then it turns out 6(6+1)=42 so you have to add 6+1 first which is 7 then multiply it by 6 which is 42 so x=6

Answer 4

x(x + 1 ) =42
x^2 +x - 42 =0
x^2+ 7x -6x -42 = 0
x(x +7 ) - 6(x + 7) = 0
(x + 7)(x - 6)=0
x+7= 0 or x-6 =0
x = -7 or x = 6

Answer 5

x(x+1)=42
becomes
x times {x + 1} = 42

But we know the Factors of 42 which have a DIFFERENCE of 1{ONE} are 7 and 6

So x= 6 and {x + 1} = 7 :- )

Answer 6

x^2 + x =42
x^2 + x -42

a=1 ,x^2 coef
b=1 ,x coef
c=-42 ,constant

roots of eqn :

( -b + (b^2-4ac)^2 )/2a , ( -b - (b^2-4ac)^2 )/2a

they are....
(-1 + 13) / 2 =6
(-1 - 13) /2 = -7

x(x+1) =42 where x =6 & x= -7 are values of x that satisfies....

Answer 7

x(x + 2) = 42
x^2 + 2x - 42 = 0
([x + 7][x - 6]) + x = 0

Answer 8

Get teacher's edition of the text, look up the answer in back.