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3 years ago

Which is the solution of the quadratic equation (4y-3)^2=72 ?

Mathematics

1 answer:

max2010maxim [7]3 years ago

7 0

Hello :
<span>(4y-3)²=72
4y-3 = </span>√72 or 4y-3 = - √72
y= (3+√72) /4 or y= (3 - √72) /4

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Answer:

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alexgriva [62]

Answer:

$P=\left(\begin{array}{ccc}-\frac{2}{3}&-\frac{2}{3}&\frac{1}{3}\\\frac{1}{\sqrt{5}}&0&\frac{2}{\sqrt{5}}\\-\frac{4}{3\sqrt{5}}&\frac{\sqrt{5}}{3}&\frac{2}{3\sqrt{5}}\end{array}\right)$

Step-by-step explanation:

It is a result that a matrix $A$ is orthogonally diagonalizable if and only if $A$ is a symmetric matrix. According with the data you provided the matrix should be

$A=\left(\begin{array}{ccc}-9&-4&2\\ -4&-9&2\\2&2&-6\\\end{array}\right)$

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With this in mind we can form the matrices $P, D$ that diagonalizes the matrix $A$ so.

$P=\left(\begin{array}{ccc}-2&-2&1\\1&0&2\\-1&1&0\\\end{array}\right)$

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$D=\left(\begin{array}{ccc}-14&0&0\\0&-5&0\\0&0&-5\\\end{array}\right)$

Observe that the rows of $P$ are the eigenvectors corresponding to the eigen values.

Now you only need to normalize each row of $P$ dividing by its norm, as a row vector.

The matrix you have to obtain is the matrix shown below

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