0
Add your answer:
Earn +20 pts
X2 plus 7x plus 6 equals 0?
(x + 6)(x + 1) = 0 so x = either -1 or -6
X2 plus 5x-6 equals 0?
x^2+5-6=0 (x+6)(x-1)=0 x=1,-6
What are the possible values for x if x squared plus 5x equals 6?
x2 + 5x =6 x2+5x -6 = 0 (x+6)(x-1) = 0 x+ 6 = 0 x = -6 x-1 = 0 x = 1
X squared-7x plus 6 equals 0?
x2 - 7x + 6 = 0(x - 6)(x - 1) = 0x = 6, x = 1
Solve the equation- X cubed plus 4x squared plus x minus 6 equals 0?
x3 + 4x2 + x - 6 = 0 x3 - x2 + 5x2 - 5x + 6x - 6 = 0 x2(x - 1) + 5x(x - 1) + 6(x - 1) = 0 (x - 1)(x2 + 5x + 6) = 0 (x - 1)(x + 2)(x + 3) = 0 So x = 1 or x = -2 or x = -3
X2 plus 7x plus 6 equals 0?
(x + 6)(x + 1) = 0 so x = either -1 or -6
X2 plus 5x-6 equals 0?
x^2+5-6=0 (x+6)(x-1)=0 x=1,-6
What are the possible values for x if x squared plus 5x equals 6?
x2 + 5x =6 x2+5x -6 = 0 (x+6)(x-1) = 0 x+ 6 = 0 x = -6 x-1 = 0 x = 1
X squared-7x plus 6 equals 0?
x2 - 7x + 6 = 0(x - 6)(x - 1) = 0x = 6, x = 1
Solve the equation- X cubed plus 4x squared plus x minus 6 equals 0?
x3 + 4x2 + x - 6 = 0 x3 - x2 + 5x2 - 5x + 6x - 6 = 0 x2(x - 1) + 5x(x - 1) + 6(x - 1) = 0 (x - 1)(x2 + 5x + 6) = 0 (x - 1)(x + 2)(x + 3) = 0 So x = 1 or x = -2 or x = -3
What are the roots of the equation x2-7x plus 6 equals 0?
x2- 7x + 6 = 0 factor, (x-6)(x-1) = 0 x = 6 x = 1
X2 plus 5x plus 3 equals 9?
x2 + 5x + 3 = 9 x2 + 5x - 6 = 0 (x + 6)(x - 1) = 0 x = -6 or x = 1
X2 plus 7x plus 9 equals 3?
x2 + 7x + 9 = 3 ∴ x2 + 7x + 6 = 0 ∴ (x + 6)(x + 1) = 0 ∴ x ∈ {-6, -1}
What are the roots of 6x2 plus 37x plus 6?
When the given expression equals 0 then x = -1/6 and x = -6
What are the values of X in the equation 2x2 equals 10x plus 12?
2x2 = 10x + 12 2x2 - 10x - 12 = 0 x2 -5x -6 = 0 x2 - 6x + x - 6 = 0 x(x -6) + 1(x -6) = 0 (x+1)(x-6) = 0 x = -1 or x = 6
X2 plus 2x plus -6 equals 0?
x2 + 2x -6 = 0 x2 + 2x + 1 = 7 (x + 1)2 = 7 x = -1 ± √7
What are the solutions to the simultaneous equations of y plus 1 equals x and y equals x square -6x plus 5?
y + 1 = xy = x² - 6x + 5Rearrange {1} to make y the subject, substitute for y in the second and solve the quadratic:y + 1 = x → y = x - 1y = x² - 6x + 5→ x - 1 = x² - 6x + 5→ x² - 7x + 6 = 0→ (x - 1)(x - 6) = 0→ x = 1 or 6→ x = 1: y = 1 - 1 = 0& x = 6 → y = 6 - 1 = 6→ the solutions are the points (1, 0) and (6, 5)
