x2 - x = 0 ?
simplifying to the initial equation by grouping of terms:
x(x-1) = 0
is true when x = 1 or x = 0
by the quadratic formula we can find the root (solution) of the equation as follows:
the equation can be rewritten as
(1(x2)) - (1(x)) + 0 = 0
so that we can see the form ax2 + bx +c = 0
Substituting our values for a, b and c into the formula, we get
x = (-b ± √(b2 - 4ac)) ÷ 2a
x = (-(-1) ± √(-12 - 4x1x0)) ÷ (2 x 1)
x = (-(-1) ± √(1 - 0)) ÷ 2
x = (1 ± 1) ÷ 2
x = 2 ÷ 2 = 1
OR
x = 0 ÷ 2 = 0
i.e. the roots (solution) of the equation x2 - x = 0 ? are when x = 1 or x = 0
They touch each other at (0, 100) on the x and y axis.
(x−6)(x+5)
x + 1
(x+4)(x-7)
x squared plus 5
x = ? 42 = x squared minus x
By factoring I get x-3 divided by x+3
-22 - -22
xsquared minus x minus 72
They touch each other at (0, 100) on the x and y axis.
(x - 5)(x + 4)
(x−6)(x+5)
Sin squared, cos squared...you removed the x in the equation.
The answer to x2 - 2x - 4y2 - 4y =(x - 2y)(x - 2y - 2)
x + 1
(x+4)(x-7)
x^(2) - 12x + 36 Look for two factors of '36', that add/subtract to '12'. ;- 1,36 ; 2,18; 4,9; 6,6. The last pair 6 & 6 add to ;12; Hence (x 6 )(x 6) Which signs???? Since '36' is positive(+), then both the signs are the same. They can be both negative , or both positive. Since the sign in '12x' is hegative then they are both negative. Hence (x - 6)(x - 6) Done!!!!!