Sometimes it's easy to write a quadratic using factors and sometimes it's too difficult.

x[sup:25jc96ij]2[/sup:25jc96ij]-2x-1 is a case where you must use the alternative, the quadratic formula.

ax[sup:25jc96ij]2[/sup:25jc96ij]+bx+c=0 for x= {-b+sqrt(b[sup:25jc96ij]2[/sup:25jc96ij]-4ac)}/2a and x = {-b-sqrt(b[sup:25jc96ij]2[/sup:25jc96ij]-4ac)}/2a.

The smaller one will be the one with the - after -b.

You can do the same with the 2nd one, but that one is easy to write factors for.

Two numbers multiply to give 8 and add to give -6.

They are -4 and -2,

so x[sup:25jc96ij]2[/sup:25jc96ij]-6x+8 = (x-4)(x-2).

If this is zero, what could x be ? since zero multiplied by anything is zero.

The two values of x that cause that to be zero are x[sub:25jc96ij]1[/sub:25jc96ij] and x[sub:25jc96ij]2[/sub:25jc96ij].

For the 3rd, you can rewrite that as x[sup:25jc96ij]2[/sup:25jc96ij] = 125.

Now 3(3) = 9, but -3(-3) is also 9, so what is the value of x[sub:25jc96ij]1[/sub:25jc96ij]?