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

Question

10245 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-19T18:02:53

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

$x^2-16x+64+0 rArr (x-8)^2=0$

$rArr x=8, &, x=8$

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

Answer 2 · 2016-08-20T21:36:25

For the quadratic equation $ax^2+bx+c=0$ , the discriminant $Delta=b^2-4ac$ tells us about the nature of the equation's roots:

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

$Delta=b^2-4ac=(-16)^2-4(1)(64)=256-256=0$

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

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