# How many solutions does the equation x^2 - 16x + 64 = 0 have?

Aug 19, 2016

Conventionally, we say that the eqn. has $2$ identical roots : 8&8.

#### Explanation:

${x}^{2} - 16 x + 64 + 0 \Rightarrow {\left(x - 8\right)}^{2} = 0$

rArr x=8, &, x=8

Conventionally, we say that the eqn. has $2$ identical roots : $x = 8 , 8$.

Aug 20, 2016

For the quadratic equation $a {x}^{2} + b x + c = 0$, the discriminant $\Delta = {b}^{2} - 4 a c$ tells us about the nature of the equation's roots:

• If $\Delta > 0$, the equation has two real roots.
• If $\Delta = 0$, the equation has a single double root.
• If $\Delta < 0$, the equation has no real solutions.

So, for ${x}^{2} - 16 x + 64 = 0$, we see that $a = 1$, $b = - 16$, and $c = 64$. Thus:

$\Delta = {b}^{2} - 4 a c = {\left(- 16\right)}^{2} - 4 \left(1\right) \left(64\right) = 256 - 256 = 0$

Since $\Delta = 0$, the equation has one single solution.