How do you describe the nature of the roots of the equation #x^2=4x-1#?

Question

2597 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-13T00:02:50

This quadratic has two distinct irrational real roots.

Given:

#x^2=4x-1#

Subtract #4x-4# from both sides to get:

#x^2-4x+4=3#

That is:

#(x-2)^2 = 3#

Hence:

#x-2 = +-sqrt(3)#

So:

#x = 2+-sqrt(3)#

So this quadratic equation has two distinct irrational real roots.

Note that instead of the full derivation of the roots, we could have examined the discriminant...

Given:

#x^2=4x-1#

Subtract #4x-1# from both sides to get:

#x^2-4x+1 = 0#

This is in standard form:

#ax^2+bx+c = 0#

with #a=1#, #b=-4# and #c=1#.

This has discriminant #Delta#, given by the formula:

#Delta = b^2-4ac = (-4)^2-4(1)(1) = 16-4 = 12 = 2^2*3#

Since #Delta > 0# we can deduce that our quadratic has two distinct real roots.

Note also that #Delta# is not a perfect square, so those roots are irrational.