Assuming the missing operator before the final 5 is subtract, then: x2 - 4x - 5 = (x + 1) (x - 5) ⇒ x = -1 or 5. If it is add, there are no real root solution. The solution to the latter case is: x2 - 4x + 5 = (x - 2 - i)(x - 2 + i) ⇒ x = 2 + i or x = 2 - i.
x2 + 10x + 21 = (x + 3)(x + 7)
(x2+10x+21)/(x+3) = (x+7)
x2+10x+21 = (x+3)(x+7) when factored.....
It is: x2-10x+21 = 0 and the value of x is 3 or 7 when solved
if x2 = 5 the value of x has to be 5 divided by 2 which is 2.5 x = 2.5
x2 + 8x - 5 = 0 ∴ x2 + 8x + 16 = 21 ∴ (x + 4)2 = 21 ∴ x + 4 = ± √21 ∴ x = -4 ± √21
x=5/2 or 2.5
Assuming the missing operator before the final 5 is subtract, then: x2 - 4x - 5 = (x + 1) (x - 5) ⇒ x = -1 or 5. If it is add, there are no real root solution. The solution to the latter case is: x2 - 4x + 5 = (x - 2 - i)(x - 2 + i) ⇒ x = 2 + i or x = 2 - i.
x2 + 10x + 21 = (x + 3)(x + 7)
(x2+10x+21)/(x+3) = (x+7)
f(x) = x2 + 3 ----> f(5) = (5)2 + 3 ----> f(5) = 28
The answer would be 3x if 'X2' is '2x', but if 'X2' is x2, then the total answer would be x2 + x.
x2 + 5x - 25 is a polynomial that can't be factored. If however it has a definite value, then the value of x can be calculated. x2 + 5x - 25 = c x2 + 5x + 25 = c - 50 (x + 5)2 = c - 50 x + 5 = ±(c - 50)1/2 x = ±(c - 50)1/2 - 5 So if c = 0, x would be equal to both 5i√2 - 5 and -(5i√2 + 5)
x2+10x+21 = (x+3)(x+7) when factored.....
It is: x2-10x+21 = 0 and the value of x is 3 or 7 when solved
The factors of x2 are x and x.