What is the quadratic formula for $f (b) = b^{2} - 4 b + 4 = 0$ $f(b)=b^2 - 4b + 4 = 0$ ?

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What is the quadratic formula for $f (b) = b^{2} - 4 b + 4 = 0$ $f(b)=b^2 - 4b + 4 = 0$ ?

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Answer 1 · 2015-10-22T13:06:58

Rewriting $f (b)$ $f(b)$ as $f (x)$ $f(x)$ will permit you to use the standard formula with less confusion (since the standard quadratic formula uses $b$ $b$ as one of its constants)

Explanation:

(since the given equation uses $b$ $b$ as a variable, we will need to express the quadratic formula, which normally uses $b$ $b$ as a constant, with some variant, $ˆ b$ $hatb$ .

To help reduce confusion, I will rewrite the given $f (b)$ $f(b)$ as
$XX f (x) = x^{2} - 4 x + 4 = 0$ $color(white)("XX")f(x)=x^2-4x+4=0$

For the general quadratic form:
$XX ˆ a x^{2} + ˆ b x + ˆ c = 0$ $color(white)("XX")hatax^2+hatbx+hatc=0$
the solution given by the quadratic equation is
$XX x = \frac{- ˆ b \pm \sqrt{{ˆ b}^{2} - 4 ˆ a ˆ c}}{2 ˆ a}$ $color(white)("XX")x=(-hatb+-sqrt(hatb^2-4hatahatc))/(2hata)$

With $ˆ a = 1$ $hata = 1$ , $ˆ b = - 4$ $hatb=-4$ , and $ˆ c = + 4$ $hatc=+4$
we get
$XX b = (x =) \frac{4 \pm \sqrt{{(- 4)}^{2} + 4 (1) (4)}}{2 (1)}$ $color(white)("XX")b=(x=)(4+-sqrt((-4)^2+4(1)(4)))/(2(1))$
as the quadratic formula

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What is the quadratic formula for $f (b) = b^{2} - 4 b + 4 = 0$ $f(b)=b^2 - 4b + 4 = 0$ ?

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Explanation:

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What is the quadratic formula for f(b)=b^2 - 4b + 4 = 0f(b)=b2−4b+4=0f(b)=b^2 - 4b + 4 = 0?

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Explanation:

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What is the quadratic formula for $f (b) = b^{2} - 4 b + 4 = 0$ $f(b)=b^2 - 4b + 4 = 0$ ?