An algebra problem by Wildan Bagus Wicaksono

Algebra Level 2

If 2017 x 2 + 2018 x + 56 + 2017 x 2 + 2018 x − 56 = 112 \sqrt { 2017{ x }^{ 2 }+2018x+56 } +\sqrt { 2017{ x }^{ 2 }+2018x-56 } =112 , determine the value of 2017 x 2 + 2018 x + 56 − 2017 x 2 + 2018 x − 56 \sqrt { 2017{ x }^{ 2 }+2018x+56 } -\sqrt { 2017{ x }^{ 2 }+2018x-56 }

1 1 3 3 3 2 \frac { 3 }{ 2 } 2 2 1 2 \frac { 1 }{ 2 }

View wiki
Wildan Bagus Wicaksono by Wildan Bagus Wicaksono , 18, Indonesia

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

3 solutions

Let 2017 x 2 + 2018 x + 56 = p 2017 x 2 + 2018 x − 56 = q \sqrt { 2017{ x }^{ 2 }+2018x+56 } =p\\ \sqrt { 2017{ x }^{ 2 }+2018x-56 } =q

112 ( p − q ) = ( p + q ) ( p − q ) 112 ( p − q ) = p 2 − q 2 112 ( p − q ) = ( 2017 x 2 + 2018 x + 56 ) 2 − ( 2017 x 2 + 2018 x − 56 ) 2 112 ( p − q ) = 2017 x 2 + 2018 x + 56 − ( 2017 x 2 + 2018 x − 56 ) 112 ( p − q ) = 2017 x 2 + 2018 x + 56 − 2017 x 2 − 2018 x + 56 112 ( p − q ) = 112 p − q = 1 112(p-q)=(p+q)(p-q)\\ 112(p-q)={ p }^{ 2 }-{ q }^{ 2 }\\ 112(p-q)={ \left( \sqrt { 2017{ x }^{ 2 }+2018x+56 } \right) }^{ 2 }-{ \left( \sqrt { 2017{ x }^{ 2 }+2018x-56 } \right) }^{ 2 }\\ 112(p-q)=2017{ x }^{ 2 }+2018x+56-(2017{ x }^{ 2 }+2018x-56)\\ 112(p-q)=2017{ x }^{ 2 }+2018x+56-2017{ x }^{ 2 }-2018x+56\\ 112(p-q)=112\\ p-q=1

So, 2017 x 2 + 2018 x + 56 − 2017 x 2 + 2018 x − 56 = 1 \sqrt { 2017{ x }^{ 2 }+2018x+56 } - \sqrt { 2017{ x }^{ 2 }+2018x-56 } = 1 .

6 Helpful 0 Interesting 1 Brilliant 0 Confused

  1. Please avoid having your notation do double duty. IE Is x x a variable, or is x = p − q x = p - q ?

  • The last line is incorrect.

  • You should still verify that there is a solution to the condition of the problem.

    • Calvin Lin Staff - 3 years, 9 months ago

    Log in to reply

    thanks. There was an error when I made the solution.

    Wildan Bagus Wicaksono - 3 years, 9 months ago
    Chew-Seong Cheong
    Aug 27, 2017

    2017 x 2 + 2018 x + 56 + 2017 x 2 + 2018 x − 56 = 112 . . . ( 1 ) \begin{aligned} \sqrt{2017x^2+2018x+56} + \sqrt{2017x^2+2018x- 56} & = 112 & ... (1) \end{aligned}

    ⟹ 2017 x 2 + 2018 x + 56 = 112 − 2017 x 2 + 2018 x − 56 . . . ( 1 a ) \implies \color{#3D99F6} \sqrt{2017x^2+2018x+56} = 112 - \sqrt{2017x^2+2018x- 56} \quad ... (1a)

    Using ( 1 a ) (1a) , we have:

    2017 x 2 + 2018 x + 56 − 2017 x 2 + 2018 x − 56 = 112 − 2 2017 x 2 + 2018 x − 56 . . . ( 2 ) \begin{aligned} {\color{#3D99F6}\sqrt{2017x^2+2018x+56}} - \sqrt{2017x^2+2018x- 56} & = \color{#D61F06} 112 - 2\sqrt{2017x^2+2018x- 56} \quad ... (2) \end{aligned}

    From ( 1 ) × ( 2 ) : (1) \times (2):

    2017 x 2 + 2018 x + 56 − ( 2017 x 2 + 2018 x − 56 ) = 112 ( 112 − 2 2017 x 2 + 2018 x − 56 ) 112 = 112 ( 112 − 2 2017 x 2 + 2018 x − 56 ) ⟹ 112 − 2 2017 x 2 + 2018 x − 56 = 1 from ( 2 ) 2017 x 2 + 2018 x + 56 − 2017 x 2 + 2018 x − 56 = 1 \begin{aligned} 2017x^2+2018x+56 - \left(2017x^2+2018x- 56 \right) & = 112 \left(112 - 2\sqrt{2017x^2+2018x- 56}\right) \\ 112 & = 112 \left(112 - 2\sqrt{2017x^2+2018x- 56}\right) \\ \implies \color{#D61F06} 112 - 2\sqrt{2017x^2+2018x- 56} & = 1 \quad \quad \small \color{#D61F06} \text{from }(2) \\ \color{#D61F06} \sqrt{2017x^2+2018x+56} - \sqrt{2017x^2+2018x- 56} & = \boxed{1} \end{aligned}

    3 Helpful 0 Interesting 0 Brilliant 0 Confused
    Genis Dude
    Aug 29, 2017

    Take reciprocal of 1st equation and rationalize the denominator

    1 Helpful 0 Interesting 0 Brilliant 0 Confused

    0 pending reports

    ×

    Problem Loading...

    Note Loading...

    Set Loading...