Determine the type and number of solutions of x^2-12x+36=0

Question

Determine the type and number of solutions of x²-12x+36=0

 

Please and thank you

Answer 1

From ax²+ bx + c  which is a quadratic equation graphing to a parabola

 

Solve for the determinant ,  the part under the square root...  b²-4ac

from the quadratic equation  x =[ -b ± √(b²-4ac)]/2a

 

if b²-4ac is positive we have 2 real roots

if b²-4ac = 0 we have 1 real root

if b²-4ac is negative we have 2 imaginary roots which are complex conjugates

 

In this case we have x²-12x+ 36

a = 1  b = -12   c =36

 

b²-4ac = (-12)²-4(1)36 = 144 - 144 = 0

 

Since the determinant = 0 we will have only one real root   (x=6 as Maria stated)

 

 

 

Answer 2

There is only one solution, x = 6.  You can find it by factoring or using the quadratic formula. I prefer to factor:

 

x² - 12x + 36 = 0

(x - 6)(x - 6) = 0

 

Since both factors are the same, we need only solve: 

 

x - 6 = 0

x = 6.

 

As far as type, I'm not sure.  I can tell you that the equation represents a parabola that opens up and touches the x-axis at x = 6. 

 

I hope this helps!