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

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

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Answer 1 · 2015-10-23T21:52:37

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

Explanation:

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

$p=(4 pm sqrt(16-4*1*4))/(2*1)=(4 pm 0)/2=2$ .

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

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

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