Question

Which equation has solutions of 6 and -6? x2 – 12x + 36 = 0 x2 + 12x – 36 = 0 x2 + 36 = 0 x2 – 36 = 0

Answer 1

Answer:

the last one

Step-by-step explanation:

Answer 2

Answer:

The equation that has solutions 6 and -6 is $x^2 - 36 = 0$

Solution:

We have to find which equation has the solutions 6 and -6.

We have been given three equations.

$x^{2}-12 x+36=0$  --- eqn 1

$x^{2}+12 x-36=0$ -- eqn 2

$x^{2}-36=0$  ---- eqn 3

The 6 and -6 to satisfy any of these equations they have to be the roots of the equation.

This means that when we substitute 6 and -6 in any of the equations and then solve them the answer on simplification should be 0.

This condition should individually be satisfied by both 6 and -6 for any one of the equations.

Now let us try and substitute 6 and -6 in eq1.

Now, substituting 6 in eq1.

62-12×6+36=0

Now we simply the equation to check is the LHS is equal to the RHS of the equation.  

LHS:

72-72=0

RHS:  0  

Since LHS=RHS it is the root of the equation.

Now we check if -6 satisfies eq1.

-62-12×-6+36=0

LHS:

72+72=144

RHS:  0

Hence LHS is not equal to RHS, -6 is not the root of eq1.

Similarly we check for eq2  

Checking for 6 and -6 we get

LHS is not equal to RHS hence this does not satisfy eq2.

Now in the same way we check for eq3

LHS=RHS for both 6 and -6 hence they are the solutions for eq3.

Hence the equation that has solutions 6 and -6 is $x^2 - 36 = 0$