Why can I not use the quadratic formula here?

Question

$x^2-14x+49$

I tried applying the quadratic formula here, but I end up with $14 ± 0$ which makes no sense.

I am supposed to factor the above mentioned expression and I thought the quadratic formula would be appropriate. Any idea why it's not working?

Answer 1

There is an error, It is a perfect square $ (x-7)^2,$ both factors are same.The roots are $ 7, 7. $

EDIT1:

You can very well use the quadratic solution formula here. Using the same, you should get $\dfrac{14±0}{2} $ but you forgot dividing by 2 , and got $ 14\pm 0 $ only, double the value of double root that you ought to have got.

Answer 2

\begin{array}{*{20}c} {x = \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}} & {{\rm{when}}} & {ax^2 + bx + c = 0} \\ \end{array}

In your case: $a=1, b=-14, c=49\;$ so you get:

$$x=\frac{-(-14)\pm \sqrt{(-14)^2 -4*1*49}}{2*1}$$

$$x=14/2=7$$.

$$f(x)=(x-7)(x-7)$$