x x with squares

Algebra Level 1

If y = 15 y = 15 , for what real value of x x do we have

x 3 + y + 4 x = 160 x^3+y + 4x = 160


The answer is 5.

Connor Switala by Connor Switala , 21, USA

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4 solutions

x^3 + 4x = 145

x(x^2+4) = 5 * 29

So, x = 5.

5 Helpful 0 Interesting 0 Brilliant 0 Confused

x 3 + 4 x = 145 x^3 + 4x = 145

x ( x 2 + 4 ) = 5 ⋅ 29 x (x^2 +4) = 5 \cdot 29

145 + y [ y = 15 ] y [y = 15] = 160 160

So, x = 5 x = 5

Connor Switala - 6 years, 8 months ago

x 3 + 4 x = 145 x^3 + 4x = 145 x ( x 2 + 4 ) = 5 ⋅ 29 x(x^2 + 4) = 5 \cdot 29 So x = 5 x=5

Connor Switala - 6 years, 7 months ago
Jade Sy
Jan 13, 2017

I solved it in a minute solution writing.

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Although some people might start solving the cubic equation,but in this case the problem is able to be solved by an easier method.Substitute y = 15 y=15 . x 3 + 15 + 4 x = 160 x 3 + 4 x = 160 − 15 = 145 x ( x 2 + 4 ) = 145 \color{#3D99F6}{x^3+15+4x=160\\x^3+4x=160-15=145\\x(x^2+4)=145} Note that 145 = 5 × 29 = 5 × ( 25 + 4 ) = 5 × ( 5 2 + 4 ) \color{#20A900}{145=5\times29=5\times(25+4)=5\times(5^2+4)} .So the equation can be re-written as: x ( x 2 + 4 ) = 5 ( 5 2 + 4 ) → x = 5 \color{#D61F06}{x(x^2+4)=5(5^2+4)\rightarrow \boxed{x=5}}

0 Helpful 0 Interesting 0 Brilliant 0 Confused

x^3+15+4x=160 x^3+4x=145. x * (x^2 +4 )= 5 * (5^2+4). so, x=5

0 Helpful 0 Interesting 0 Brilliant 0 Confused

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