Squares and square roots: Equations
Find the real value(s) of x that verify: βπ₯ + β7π₯ + 2 = 0
solution
Answer: NO REAL VALUES.
Work: Let βπ₯ be a and β (7x +2) be b
If a>=0, b>=0 thenβπ₯ (a) + β (7x +2)(b)=0,
Then a=b=0, for this equation to be defined, x must be >=0.
We solve for βπ₯ =0 we get x=0,
7x+2 = 0, x= -2
7
This means x = 0 and x =-2/7Β which is impossible. Hence the equation has no real values.