How do you solve $x^{3} + 147 = 3 x^{2} + 49 x$ $x^3 + 147 = 3x^2 + 49x$ ?

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How do you solve $x^{3} + 147 = 3 x^{2} + 49 x$ $x^3 + 147 = 3x^2 + 49x$ ?

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Answer 1 · 2015-11-27T22:43:56

Subtract $147 + 49 x$ $147+49x$ from both sides, then identify common factors on each side to find:

$x = 3$ $x = 3$ , $7$ $7$ or $- 7$ $-7$

Explanation:

Subtract $147 + 49 x$ $147+49x$ from both sides to get:

$x^{3} - 49 x = 3 x^{2} - 147$ $x^3-49x = 3x^2-147$

That is:

$x (x^{2} - 49) = 3 (x^{2} - 49)$ $x(x^2-49) = 3(x^2-49)$

So either $x = 3$ $x = 3$ or $x^{2} - 49 = 0$ $x^2-49 = 0$ , giving $x = \pm \sqrt{49} = \pm 7$ $x = +-sqrt(49) = +-7$

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Alternatively and more systematically:

Subtract the right hand side from the left to get:

$x^{3} - 3 x^{2} - 49 x + 147 = 0$ $x^3-3x^2-49x+147 = 0$

Let $f (x) = x^{3} - 3 x^{2} - 49 x + 147$ $f(x) = x^3-3x^2-49x+147$

Factor by grouping:

$x^{3} - 3 x^{2} - 49 x + 147$ $x^3-3x^2-49x+147$

$= (x^{3} - 3 x^{2}) - (49 x - 147)$ $=(x^3-3x^2)-(49x-147)$

$= x^{2} (x - 3) - 49 (x - 3)$ $=x^2(x-3)-49(x-3)$

$= (x^{2} - 49) (x - 3)$ $=(x^2-49)(x-3)$

$= (x^{2} - 7^{2}) (x - 3)$ $=(x^2-7^2)(x-3)$

$= (x - 7) (x + 7) (x - 3)$ $=(x-7)(x+7)(x-3)$

...using the difference of squares identity:

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

So $f (x) = 0$ $f(x) = 0$ has roots $x = 7$ $x = 7$ , $x = - 7$ $x = -7$ and $x = 3$ $x = 3$

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How do you solve $x^{3} + 147 = 3 x^{2} + 49 x$ $x^3 + 147 = 3x^2 + 49x$ ?

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Explanation:

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How do you solve x3+147=3x2+49xx^3 + 147 = 3x^2 + 49x?

1 Answer

Explanation:

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How do you solve $x^{3} + 147 = 3 x^{2} + 49 x$ $x^3 + 147 = 3x^2 + 49x$ ?