x2 + 10x = 0 x2 + 10x + 25 = 25 (x + 5)2 = 25 x + 5 = +-5 x1 = 0 x2 =10
2x2=4 x2=8 x2=16 x2 = 32 and 5x5 = 25 so no, I guess not.
x2 - 10x + 25 = 0(x - 5)(x - 5) = 0x - 5 = 0x = 5
x2 + 10x = 18 x2 + 2(5)x - 18 = 0 x2 + 2(5)x + 25 = 18 + 25 (x + 5) 2 = 43 x + 5 = 430.5 x=6.56(3sf)-5
25
No, not always. It depends on if the original biconditional statement is true. For example take the following biconditional statement:x = 3 if and only if x2 = 9.From this biconditional statement we can extract two conditional statements (hence why it is called a bicondional statement):The Conditional Statement: If x = 3 then x2 = 9.This statement is true. However, the second statement we can extract is called the converse.The Converse: If x2=9 then x = 3.This statement is false, because x could also equal -3. Since this is false, it makes the entire original biconditional statement false.All it takes to prove that a statement is false is one counterexample.
x2 = -25 There exists no real number for which this is true, so we must use imaginary numbers... x = 5i
x2 + 10x + 25 --> (x + 5)2
x2 + 10x = 0 x2 + 10x + 25 = 25 (x + 5)2 = 25 x + 5 = +-5 x1 = 0 x2 =10
x2 - 6x = 16 ∴ x2 - 6x + 9 = 25 ∴ (x - 3)2 = 25 ∴ x - 3 = 25 ∴ x = 28
x2+4x+4 = 25 x2+4x+4-25 = 0 x2+4x-21 = 0 (x+7)(x-3) = 0 x = -7 or x = 3
If x2 < 25 Then: |x| < 5 -5 < x < 5
x^2<25
In ordinary mathematics, assuming that x = X and that X2 denotes x2 or x-squared, there cannot be a counterexample since the statement is TRUE. However, there are two assumptions made that could be false and so could give rise to counterexamples. 1. x is not the same as X. If, for example X = 4x then X = -20 so that X2 = 400. 2a. X2 is not X2 but X times 2. In that case X2 = -10. 2b. X2 is x2 modulo 7, for example. Then X2 = 4.
5. A circle with centre (0,0) has equation: x2 + y2 = radius2 With: x2 + y2 = 25 = 52 The radius is 5.
2x2=4 x2=8 x2=16 x2 = 32 and 5x5 = 25 so no, I guess not.
Rearrange: x2 - 10x + 25 = 0ie (x - 5)(x - 5) = 0x = 5Check: x2 = 25; 10x - 25 = 50 - 25 = 25. Job done