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3 years ago

How do I solve for x in x^2 - 8x + 12 = 0 using factoring?

Mathematics

2 answers:

jonny [76]3 years ago

7 0

$x^2 - 8x + 12 = 0\\
x^2-2x-6x+12=0\\
x(x-2)-6(x-2)=0\\
(x-6)(x-2)=0\\
x=6 \vee x=2$

gladu [14]3 years ago

4 0

$\rm x^2-8x+12=0 \\ \\ x^2-2x-6x+12=0 \\ \\ x(x-2)-6(x-2)=0 \\ \\ (x-2)(x-6)=0 \\ \\ \bold{I:x-2=0 \longrightarrow x=2} \\ \bold{II:x-6=0 \longrightarrow x=6}$

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Answer:

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Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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This means that when X = 15.5, Z has a pvalue of 0.07. So when X = 15.5, Z = -1.475.

$Z = \frac{X - \mu}{\sigma}$

$-1.475 = \frac{15.5 - \mu}{\sigma}$

$15.5 - \mu = -1.475\sigma$

$\mu = 15.5 + 1.475\sigma$

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This means that when X = 16.3, Z has a pvalue of 1-0.1 = 0.9. So when X = 16.3, Z = 1.28.

$Z = \frac{X - \mu}{\sigma}$

$1.28 = \frac{16.3 - \mu}{\sigma}$

$16.3 - \mu = 1.28\sigma$

$\mu = 16.3 – 1.28\sigma$

From above

$\mu = 15.5 + 1.475\sigma$

$15.5 + 1.475\sigma = 16.3 – 1.28\sigma$

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$\sigma = \frac{0.8}{2.755}$

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