How do you find all the zeros of $x^{2} - 8 x - 48 = 0$ $x^2 - 8x - 48 = 0$ with its multiplicities?

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How do you find all the zeros of $x^{2} - 8 x - 48 = 0$ $x^2 - 8x - 48 = 0$ with its multiplicities?

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Answer 1 · 2016-04-16T19:41:01

$x = 12$ $x=12$ or $x = - 4$ $x=-4$ , each with multiplicity $1$ $1$

Explanation:

Complete the square then use the difference of squares identity:

$a^{2} - b^{2} = (a - b) (a + b)$ $a^2-b^2 = (a-b)(a+b)$

with $a = (x - 4)$ $a=(x-4)$ and $b = 8$ $b=8$ as follows:

$0 = x^{2} - 8 x - 48$ $0 = x^2-8x-48$

$= {(x - 4)}^{2} - 16 - 48$ $=(x-4)^2-16-48$

$= {(x - 4)}^{2} - 8^{2}$ $=(x-4)^2-8^2$

$= ((x - 4) - 8) ((x - 4) + 8)$ $=((x-4)-8)((x-4)+8)$

$= (x - 12) (x + 4)$ $=(x-12)(x+4)$

So $x = 12$ $x=12$ or $x = - 4$ $x=-4$ , each with multiplicity $1$ $1$

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How do you find all the zeros of $x^{2} - 8 x - 48 = 0$ $x^2 - 8x - 48 = 0$ with its multiplicities?

1 Answer

Explanation:

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Science

Math

Humanities

... and beyond

How do you find all the zeros of x2−8x−48=0x^2 - 8x - 48 = 0 with its multiplicities?

1 Answer

Explanation:

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How do you find all the zeros of $x^{2} - 8 x - 48 = 0$ $x^2 - 8x - 48 = 0$ with its multiplicities?