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2 years ago

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Solve the following quadratic equation. (x+12)^2=1 A. x = 11 and x = 13 B. x = -11 and x = -13 C. x = -11 and x = 13 D. x = 11 a

nd x = -13
Will make brainiest!!!

1 answer:

sveticcg [70]2 years ago

7 0

Answer:

b

Step-by-step explanation:

(x+12)^2=1

or

(x+12)= +or - 1

when

x+12=1

x=1-12 =-11

when

x+12=-1

x=-1-12 =-13

then

x=-11 and x= -13

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P(x) = x + 1x² – 34x + 343<br> d(x)= x + 9

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$x=\frac{9}{d-1},\:P=\frac{-297d+378}{\left(d-1\right)^2}+343$

Step-by-step explanation:

Let us start by isolating x for dx = x + 9.

dx - x = x + 9 - x > dx - x = 9.

Factor out the common term of x > x(d - 1) = 9.

Now divide both sides by d - 1 > $\frac{x\left(d-1\right)}{d-1}=\frac{9}{d-1};\quad \:d\ne \:1$ . Go ahead and simplify.

$x=\frac{9}{d-1};\quad \:d\ne \:1$ .

Now, $\mathrm{For\:}P=x+1x^2-34x+343, \mathrm{Subsititute\:}x=\frac{9}{d-1}$ .

$P=\frac{9}{d-1}+1\cdot \left(\frac{9}{d-1}\right)^2-34\cdot \frac{9}{d-1}+343$ .

Group the like terms... $1\cdot \left(\frac{9}{d-1}\right)^2+\frac{9}{d-1}-34\cdot \frac{9}{d-1}+343$ .

$\mathrm{Add\:similar\:elements:}\:\frac{9}{d-1}-34\cdot \frac{9}{d-1}=-33\cdot \frac{9}{d-1}$ > $1\cdot \left(\frac{9}{d-1}\right)^2-33\cdot \frac{9}{d-1}+343$ .

Now for $1\cdot \left(\frac{9}{d-1}\right)^2 > \mathrm{Apply\:exponent\:rule}: \left(\frac{a}{b}\right)^c=\frac{a^c}{b^c} > \frac{9^2}{\left(d-1\right)^2} = 1\cdot \frac{9^2}{\left(d-1\right)^2}$ .

$\mathrm{Multiply:}\:1\cdot \frac{9^2}{\left(d-1\right)^2}=\frac{9^2}{\left(d-1\right)^2}$ .

Now for $33\cdot \frac{9}{d-1} > \mathrm{Multiply\:fractions}: \:a\cdot \frac{b}{c}=\frac{a\:\cdot \:b}{c} > \frac{9\cdot \:33}{d-1} > \frac{297}{d-1}$ .

Thus we then get $\frac{9^2}{\left(d-1\right)^2}-\frac{297}{d-1}+343$ .

Now we want to combine fractions. $\frac{9^2}{\left(d-1\right)^2}-\frac{297}{d-1}$ .

$\mathrm{Compute\:an\:expression\:comprised\:of\:factors\:that\:appear\:either\:in\:}\left(d-1\right)^2\mathrm{\:or\:}d-1 > This\: is \:the\:LCM > \left(d-1\right)^2$

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$\frac{9^2-297\left(d-1\right)}{\left(d-1\right)^2} > 9^2=81 > \frac{81-297\left(d-1\right)}{\left(d-1\right)^2}$ .

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$-297d-\left(-297\right)\cdot \:1 > \mathrm{Apply\:minus-plus\:rules} > -\left(-a\right)=a > -297d+297\cdot \:1$ .

$\mathrm{Multiply\:the\:numbers:}\:297\cdot \:1=297 > -297d+297 > 81-297d+297 > \mathrm{Add\:the\:numbers:}\:81+297=378 > -297d+378 > \frac{-297d+378}{\left(d-1\right)^2}$

Therefore $P=\frac{-297d+378}{\left(d-1\right)^2}+343$ .

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