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3 years ago

Solve the equation by Completing the Square x2 – 10x + 13 = 0

Mathematics

2 answers:

kifflom [539]3 years ago

7 0

X^2 -10x = -13
c= ( -10/2)^2
= (-5)^2
= 25

x^2 -10x+25= -13+25
x^2-10x+25= 12
(x-5) (x-5) =12
x-5 =12
x-5+5=12+5
x= 17

user100 [1]3 years ago

4 0

Answer:

$x=5+\sqrt{12}$

$x=5-\sqrt{12}$

Step-by-step explanation:

$x^2-10x+13=0$

$x^2-10x=-13$

$x^2-10x+(\frac{10}{2})^2 =-13+(\frac{10}{2})^2$

$x^2-10x+(5)^2=-13+(5)^2$

$x^2-10x+25=-13+25\\x^2-10x+25=12$

Factor.

$(x-5)^2=12$

Extract the square root.

$x-5=\frac{+}{} \sqrt{12}$

Add 5

$x=5+\sqrt{12}$

$x=5-\sqrt{12}$

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Answer:

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Step-by-step explanation:

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Answer:

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The, we case correspond to find the coordinates of vectors of C,

$\{\left[\begin{array}{ccc}2\\-1\\-1\end{array}\right], \left[\begin{array}{ccc}2\\0\\-1\end{array}\right], \left[\begin{array}{ccc}-3\\1\\2\end{array}\right] \}$

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1. We need to find $a,b,c\in\mathbb{R}$ such that

$\left[\begin{array}{ccc}2\\-1\\-1\end{array}\right]=a\left[\begin{array}{ccc}1\\-1\\0\end{array}\right]+b\left[\begin{array}{ccc}-2\\2\\-1\end{array}\right]+c\left[\begin{array}{ccc}2\\-1\\1\end{array}\right]$

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$\left[\begin{array}{cccc}1&-2&2&2\\0&-1&1&-1\\0&0&1&1\end{array}\right]$

now we use backward substitution

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2. We need to find $a,b,c\in\mathbb{R}$ such that

$\left[\begin{array}{ccc}2\\0\\-1\end{array}\right]=a\left[\begin{array}{ccc}1\\-1\\0\end{array}\right]+b\left[\begin{array}{ccc}-2\\2\\-1\end{array}\right]+c\left[\begin{array}{ccc}2\\-1\\1\end{array}\right]$

Then we find these values solving the linear system

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$\left[\begin{array}{cccc}1&-2&2&2\\0&-1&1&-1\\0&0&1&2\end{array}\right]$

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$\left[\begin{array}{ccc}-3\\1\\2\end{array}\right]=a\left[\begin{array}{ccc}1\\-1\\0\end{array}\right]+b\left[\begin{array}{ccc}-2\\2\\-1\end{array}\right]+c\left[\begin{array}{ccc}2\\-1\\1\end{array}\right]$

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