Question

1)√x^6=-x^3 2) √c^4=-c^2

Answer 1

√х∧6=1/х∧3⇒(√х∧6 )×х∧3=1

(√х∧4×х∧2) ×х∧3=1

х∧2×х×х∧3=1

х∧6=1

х=1

2)√с∧4=1/с∧2

(√с∧2×с∧2) ×с∧2=1

с×с×с∧2=1

с∧4=1

с=1

Answer 2

Let's solve each equation step-by-step:

1. √x^6 = -x^3

To solve this equation, we'll get rid of the square root by squaring both sides of the equation:

(√x^6)^2 = (-x^3)^2

x^6 = x^6

Now, the equation simplifies to:

x^6 - x^6 = 0

Since both terms on the left side are the same with opposite signs, they cancel each other out, leaving us with:

0 = 0

This equation is true for all real values of x. In other words, the original equation has an infinite number of solutions, which means any real value of x satisfies the equation.

2. √c^4 = -c^2

Again, we'll remove the square root by squaring both sides of the equation:

(√c^4)^2 = (-c^2)^2

c^4 = c^4

Now, the equation simplifies to:

c^4 - c^4 = 0

Just like in the first equation, both terms on the left side are the same with opposite signs, so they cancel each other out:

0 = 0

As with the first equation, this equation is true for all real values of c. Therefore, any real value of c satisfies this equation as well.

In summary, both equations have an infinite number of solutions, and any real value of the respective variable (x or c) will satisfy the equations.