x2-5-4x2+3x = 0 -3x2+3x-5 = 0 or as 3x2-3x+5 = 0
2x^2 + 8x + 3 = 0
I am assuming that in this question x2 refers to x2 since otherwise the second term "x2" makes no sense. Then 3x2/x2 = 3.
x3 + ax + 3a + 3x2 = x (x2 + a) + 3 (a + x2) = x (x2 + a) + 3 (x2 + a) = (x2 + a)(x + 3) Checking the work: x3 + ax + 3x2 + 3a or x3 + 3x2 + 3a + ax = x2 (x + 3) + a (3 + x) = x2 (x + 3) + a (x + 3) = (x + 3)(x2 + a)
3x2 + 27x = 30 ∴ x2 + 9x - 10 = 0 ∴ (x + 10)(x - 1) = 0 ∴ x ∈ {-10, 1}
It is: 3x2-5x-2 = 0 and the value of x is -1/3 or 2 when solved
In general, no.
x2-5-4x2+3x = 0 -3x2+3x-5 = 0 or as 3x2-3x+5 = 0
Assuming the 2 is meant to be a square this is the form: x2 - 3x2 = -2x2 + 8x + 5
2x^2 + 8x + 3 = 0
I am assuming that in this question x2 refers to x2 since otherwise the second term "x2" makes no sense. Then 3x2/x2 = 3.
0
x2+8x = -3x2+5 4x2+8x-5 = 0 (2x+5)(2x-1) = 0 x = -5/2 or x = 1/2
x3 + ax + 3a + 3x2 = x (x2 + a) + 3 (a + x2) = x (x2 + a) + 3 (x2 + a) = (x2 + a)(x + 3) Checking the work: x3 + ax + 3x2 + 3a or x3 + 3x2 + 3a + ax = x2 (x + 3) + a (3 + x) = x2 (x + 3) + a (x + 3) = (x + 3)(x2 + a)
3x2 + 27x = 30 ∴ x2 + 9x - 10 = 0 ∴ (x + 10)(x - 1) = 0 ∴ x ∈ {-10, 1}
X = 5 or -5 3x2 - 42 = 33 3x2 = 75 x2=25
3x2 = 21 describes two single values, and does not have an x intercept. 3x2 = 21 ∴ x2 = 7 ∴ x = ±√7