9x-53=6/x 9x2-53x=6 9x2-53x-6=0 9x2+x-54x-6=0 x(9x+1)-6(9x+1)=0 (9x+1)x(x-6)=0 9x+1=0 or x-6=0 9x=-1 or x=6 x=-1/9 or x=6
If you mean y =9x^2 -4 , than the range is the possible y values. Range = 0<= y < infinity.
9x2 + 18x - 16 = 0 ⇒ (3x - 2)(3x + 8) = 0 ⇒ 3x - 2 = 0 → x = 2/3 or 3x + 8 = 0 → x = -22/3
Presumably this is a quadratic equation question in the form of: 9x2-12x+4 = 0 When factored: (3x-2)(3x-2) = 0 Solution: x = 2/3 and also x = 2/3 (they both have equal roots)
Solve this problem -x squared -40x- 80 =0
9x2 + 7x = 2 9x2 + 7x - 2 = 0 9x2 + 9x - 2x - 2 = 0 9x(x + 1) - 2(x + 1) = 0 (9x - 2)(x + 1) = 0 x ∈ {-1, 2/9}
9x-53=6/x 9x2-53x=6 9x2-53x-6=0 9x2+x-54x-6=0 x(9x+1)-6(9x+1)=0 (9x+1)x(x-6)=0 9x+1=0 or x-6=0 9x=-1 or x=6 x=-1/9 or x=6
k = -3; factors are (9x + 3)(x - 1)
3x2-9x = 0 x(3x-9) = 0 x = 0 or x = 3
10x=x 9x=0 x=0
x - 10x = 9x, so 9x + 3 = 0, so 9x = -3 so x = -1/3
x2 - 9x = -8 ∴ x2 - 9x + 8 = 0 ∴ (x - 1)(x - 8) = 0 You can then solve it as well: ∴ x ∈ {1, 8}
2x2 - 9x = 18 2x2 - 9x -18 = 0 (2x - 18)(x + 9) x= -9 and 9
(3x+4)(3x-4)=0 x=±4/3
9x2 = 2x Get the 2x to the left side to get 9x2 - 2x = 0 Use cuadratic formula to solve for x knowing that c is 0 x = [ -b +- sqrt (b2-4ac) ]/2a
If you mean y =9x^2 -4 , than the range is the possible y values. Range = 0<= y < infinity.
9x2 + 18x - 16 = 0 ⇒ (3x - 2)(3x + 8) = 0 ⇒ 3x - 2 = 0 → x = 2/3 or 3x + 8 = 0 → x = -22/3