how to factorise x^2+x=42?

Answer 1

It's been about 12 years since I ever had to do this...but I think this is what you're trying to figure out:

x^2+x = 42
x^2+x-42 = 0
(x-6)(x+7) = 0, therefore
x=6, x=-7

Answer 2

First: move the 42 to the Left Hand Side (LHS) It becomes X^2+X-42=0
then u prepare two brackets : ( ) ( )=0
fill in the bracket with (x + 7 ) (x - 6 ) =0
* x times x equals to x^2
7 times -6 equals to -42
the (first x times -6) + the (second x times 7) equals to x which is the same as the number in the center of the equation

Let (x+7)=0 (x-6)=0
then x = -7 x = 6

Answer 3

first bring it in this form

x^2 + x - 42 = 0

take the coeff of x : +1

take the product of coeff of x^2 & constant : 1 * -42 = -42


now try to think of two numbers which when multiplied gives -42

& when added gives +1

-42 = -6 * 7
= 6 * -7
= 1 * -42
= -1 * 42
= 3 * -14
= -3 * 14
= 2 * -21
= -21 * 2

of all these possibilities only -6 , 7 will add to +1

so write the eqn as below

x^2 -6x + 7X -42

= x(x-6) +7(x-6)
=(x+7)(x-6)

Answer 4

x^2+x = 42
x^2+x-42 = 0
x^2+7x+6x-42 = 0
(x-6)(x+7) = 0, therefore
x=6, x=-7

Answer 5

X^2+X=42
X^2+x-42=0
X^2+(7X-6X)+(+7*-6)=0 [Factors of -42 which differ by 1 are -6 and 7]
X(X+7)-6(X+7)=0
Answer is (X+7)(X-6)

Answer 6

x² + x - 42 = 0

(x + 7)(x - 6) = 0

x = -7, 6

Answer 7

x^2+x-42=0

(x+7) (x-6)

x+7=0 or x-6=0

x=(-7) or x=6

Answer 8

Subtract 42 from both sides. SO you have zero on one side. Apply the quadratic formula and Disco!