Unnecessarily High Powered

Algebra Level 4

Suppose α \alpha is one of the roots of x 2 + x − 1 4 x^2+x-\frac{1}{4} . Determine

α 3 − 1 α 5 + α 4 − α 3 − α 2 \frac{\alpha^3-1}{\alpha^5+\alpha^4-\alpha^3-\alpha^2}


The answer is 20.

View wiki
Tom Zhou by Tom Zhou , 21, Canada

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.

4 solutions

Sandeep Rathod
Feb 13, 2015

α 3 − 1 α 5 − α 3 + α 4 − α 2 \dfrac{\alpha^3 - 1}{ \alpha^5 - \alpha^3 + \alpha^4 - \alpha^2}

α 3 − 1 α 3 ( α 2 − 1 ) + α 2 ( α 2 − 1 ) \dfrac{\alpha^3 - 1}{ \alpha^3(\alpha^2 - 1) + \alpha^2(\alpha^2 - 1)}

( α − 1 ) ( α 2 + α + 1 ) ( α 2 − 1 ) ( α 3 + α 2 ) \dfrac{(\alpha - 1)(\alpha^2 + \alpha + 1)}{(\alpha^2 - 1)( \alpha^3 + \alpha^2)}

α 2 + α + 1 ( α + 1 ) 2 α 2 \dfrac{\alpha^2 + \alpha + 1}{( \alpha + 1)^2\alpha^2}

Now,

x 2 + x − 1 4 = 0 x^2 + x - \dfrac{1}{4} = 0

⟹ α 2 + α + 1 = 5 4 \implies \alpha^2 + \alpha + 1 = \dfrac{5}{4}

x 2 + x − 1 4 = 0 x^2 + x - \dfrac{1}{4} = 0

⟹ α ( α + 1 ) = 1 4 \implies \alpha(\alpha + 1) = \dfrac{1}{4}

Thus ,

α 2 + α + 1 ( α + 1 ) 2 α 2 \dfrac{\alpha^2 + \alpha + 1}{( \alpha + 1)^2\alpha^2}

= 5 4 ( 1 4 ) 2 = \dfrac{\dfrac{5}{4}}{\left( \dfrac{1}{4}\right)^2}

= 20 = 20

78 Helpful 0 Interesting 0 Brilliant 0 Confused

Good solution..

B.s. Ashwin - 6 years, 3 months ago

I didn't got it... how ( α − 1 ) ( α 2 + α + 1 ) ( α 2 − 1 ) ( α 3 + α 2 ) = α 2 + α + 1 ( α + 1 ) 2 α 2 \dfrac{(\alpha - 1)(\alpha^2 + \alpha + 1)}{(\alpha^2 - 1)( \alpha^3 + \alpha^2)} = \dfrac{\alpha^2 + \alpha + 1}{( \alpha + 1)^2\alpha^2}

Rishabh Tripathi - 6 years, 3 months ago

Log in to reply

Factorize the denominator as alpha not equal to zero, alpha-1 will get cancelled therefore you get. :)

B.s. Ashwin - 6 years, 3 months ago

Log in to reply

Thank you soo much... i just got it right now.. ^_^

Rishabh Tripathi - 6 years, 3 months ago

( a − 1 ) ( a 2 + a + 1 ) ( a 2 − 1 ) ( a 3 + a 2 ) = ( a − 1 ) ( a 2 + a + 1 ) ( a 2 − 1 2 ) { a 2 ( a + 1 ) } ( a − 1 ) ( a 2 + a + 1 ) ( a − 1 ) ( a + 1 ) a 2 ( a + 1 ) ⟹ ( a 2 + a + 1 ) a 2 ( a + 1 ) 2 \frac { (a-1)({ a }^{ 2 }+a+1) }{ ({ a }^{ 2 }-1)({ a }^{ 3 }+{ a }^{ 2 }) } =\frac { (a-1)({ a }^{ 2 }+a+1) }{ ({ a }^{ 2 }-{ 1 }^{ 2 })\{ { a }^{ 2 }(a+1)\} } \\ \frac { (a-1)({ a }^{ 2 }+a+1) }{ (a-1)(a+1){ a }^{ 2 }(a+1) } \Longrightarrow \frac { ({ a }^{ 2 }+a+1) }{ { a }^{ 2 }{ (a+1) }^{ 2 } }

Syed Baqir - 5 years, 9 months ago

I did in the same way :)

Ahmed Arup Shihab - 6 years, 4 months ago

Nice one..

Sriram Vudayagiri - 6 years, 3 months ago
Chew-Seong Cheong
Feb 13, 2015

As α \alpha is a root of x 2 + x − 1 4 ⇒ α 2 + α − 1 4 = 0 ⇒ α 2 + α = 1 4 x^2+x-\frac{1}{4}\quad \Rightarrow \alpha^2 + \alpha - \frac {1}{4} = 0 \quad \Rightarrow \alpha^2 + \alpha = \frac {1}{4}

α 3 − 1 α 5 + α 4 − α 3 − α 2 = ( α − 1 ) ( α 2 + α + 1 ) α 3 ( α 2 + α ) − α ( α 2 + α ) = ( α − 1 ) ( 1 4 + 1 ) α 3 ( 1 4 ) − α ( 1 4 ) \dfrac {\alpha^3 - 1} {\alpha^5+\alpha^4-\alpha^3-\alpha^2} = \dfrac {(\alpha - 1)(\alpha^2+\alpha+1)} {\alpha^3 (\alpha^2+\alpha) -\alpha (\alpha^2+\alpha)} = \dfrac {(\alpha - 1)(\frac{1}{4}+1)} {\alpha^3 (\frac{1}{4}) -\alpha (\frac{1}{4})}

= 5 4 ( α − 1 ) 1 4 α ( α 2 − 1 ) = 5 ( α − 1 ) α ( α − 1 ) ( α + 1 ) = 5 α 2 + α = 5 1 4 = 20 =\dfrac {\frac {5}{4}(\alpha - 1)} {\frac{1}{4}\alpha (\alpha^2 -1)} = \dfrac {5 (\alpha - 1)} {\alpha (\alpha -1)(\alpha+1)} = \dfrac {5}{\alpha^2 + \alpha} = \dfrac {5}{\frac{1}{4}} = \boxed{20}

36 Helpful 0 Interesting 0 Brilliant 0 Confused

x 2 + x = 1 4 ∴ α 2 + α = 1 4 . . . . . . ( 1 ) α 3 − 1 α 5 − α 3 + α 4 − α 2 ∴ ( α − 1 ) ( α 2 + α + 1 ) α 3 ( α 2 − 1 ) + α 2 ( α 2 − 1 ) α ≠ 1 ∴ Dividing numerator and denominator by ( α − 1 ) a n d u s i n g ( 1 ) 1 4 + 1 α 3 ( α + 1 ) + α 2 ( α + 1 ) 5 4 ( α 2 + α ) ( α 2 + α ) F r o m ( 1 ) 5 4 ( 1 4 ) ( 1 4 ) = 20 x^2+x=\frac{1}{4}~~~~\therefore \alpha^2+\alpha=\frac{1}{4}......(1)\\\dfrac{\alpha^3 - 1}{ \alpha^5 - \alpha^3 + \alpha^4 - \alpha^2}\\\therefore \dfrac{(\alpha - 1)(\alpha^2 + \alpha + 1)}{\alpha^3(\alpha^2 - 1) +\alpha^2(\alpha^2 - 1)} \\ \alpha \ne 1 \therefore \text{ Dividing numerator and denominator by } (\alpha - 1) ~and~ using~ (1) \\ \dfrac{\frac{1}{4}+1}{\alpha^3(\alpha+ 1) +\alpha^2(\alpha+ 1)} \\ \dfrac{ \frac{5}{4}}{(\alpha^2 +\alpha)(\alpha^2+\alpha)} \\From~~(1)~~\dfrac{ \frac{5}{4} } {(\frac{1}{4})(\frac{1}{4})} =~~~\boxed{ \color{#EC7300}{ \Large 20} } \\

7 Helpful 0 Interesting 0 Brilliant 0 Confused
Curtis Clement
Feb 14, 2015

x 3 − 1 x 5 + x 4 − x 3 − x 2 = ( x − 1 ) ( x 2 + x + 1 ) x 2 ( x 3 + x 2 − x − 1 ) \dfrac{x^3 -1}{x^5 +x^4 -x^3 -x^2} = \dfrac{(x-1)(x^2+x+1)}{x^2(x^3 +x^2 -x-1)} (using the factor theorem)... = ( x − 1 ) ( x 2 + x + 1 ) x 2 ( x − 1 ) ( x + 1 ) 2 = x 2 + x + 1 x 2 ( x 2 + 2 x + 1 ) ( 1 ) \dfrac{(x-1)(x^2 +x+1)}{x^2(x-1)(x+1)^2} = \dfrac{x^2 + x +1}{x^2(x^2 +2x+1)}\ (1) Now x 2 + x − 1 4 = 0 ⇒ x 2 + x + 1 = 5 4 , x 2 = 1 4 − x x^2 + x - \frac{1}{4} = 0 \Rightarrow\ x^2 + x + 1 = \frac{5}{4} \ , \ x^2 = \frac{1}{4} - x a n d x 2 + 2 x + 1 = 5 4 + x \ and \ x^2 + 2x + 1 = \frac{5}{4} +x Substituting these into equation (1): 5 / 4 ( 1 4 − x ) ( 5 4 + x ) = 5 / 4 5 16 − x − x 2 \frac{5/4}{(\frac{1}{4} -x)(\frac{5}{4} + x)} = \frac{5/4}{\frac{5}{16} - x - x^2} x 2 + x − 1 4 = 0 ⇒ 5 16 − x − x 2 = 1 16 x^2 + x - \frac{1}{4} = 0 \Rightarrow\frac{5}{16} - x - x^2 = \frac{1}{16} ∴ 5 / 4 5 16 − x − x 2 = 5 / 4 1 / 16 = 20 \therefore\frac{5/4}{\frac{5}{16} - x - x^2} = \frac{5/4}{1/16} = 20

5 Helpful 0 Interesting 0 Brilliant 0 Confused

Feels like a circuitous route.

Richard Desper - 4 years, 5 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...