How do you solve $p^{2} - 4 p + 4 = 0$ $p^2 - 4p + 4 = 0$ using the quadratic formula?

Question

How do you solve $p^{2} - 4 p + 4 = 0$ $p^2 - 4p + 4 = 0$ using the quadratic formula?

1 Answer

Impact of this question

4940 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-10-23T21:52:37

The unique (double-root) in this case is $p = 2$ $p=2$ .

Explanation:

For an equation of the form $a x^{2} + b x + c$ $ax^2+bx+c$ , the quadratic formula says the solutions are $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x=(-b pm sqrt(b^2-4ac))/(2a)$ . For this problem, $a = 1$ $a=1$ , $b = - 4$ $b=-4$ , and $c = 4$ $c=4$ . Thus, there's just one answer:

$p = \frac{4 \pm \sqrt{16 - 4 \cdot 1 \cdot 4}}{2 \cdot 1} = \frac{4 \pm 0}{2} = 2$ $p=(4 pm sqrt(16-4*1*4))/(2*1)=(4 pm 0)/2=2$ .

Science

Math

Humanities

... and beyond

How do you solve $p^{2} - 4 p + 4 = 0$ $p^2 - 4p + 4 = 0$ using the quadratic formula?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you solve p2−4p+4=0p^2 - 4p + 4 = 0 using the quadratic formula?

1 Answer

Explanation:

Related questions

Impact of this question

How do you solve $p^{2} - 4 p + 4 = 0$ $p^2 - 4p + 4 = 0$ using the quadratic formula?