Problem 2

Algebra Level 4

$\large { \left| x \right| }^{ \frac { 6 }{ 5 } }-26{ \left| x \right| }^{ \frac { 3 }{ 5 } }-27=0$ Product of real roots of the equation above can be represented as ${ -A }^{ B }$ , where $A$ is a prime number.

Find the value of $A+B$ .

For more problems click here

The answer is 13.

2 solutions

Utkarsh Bansal
Mar 3, 2015

Put ${ \left| x \right| }^{ \frac { 3 }{ 5 } }=y$ , so that the equation can be written as ${ y }^{ 2 }-26y-27=0$ $\Rightarrow \left( y+1 \right) \left( y-27 \right) =0\quad \Rightarrow y=-1\quad or\quad 27$

As $y={ \left| x \right| }^{ \frac { 3 }{ 5 } }\ge 0\quad \forall \quad x$ , we get $y=27$

$\Rightarrow \quad { \left| x \right| }^{ \frac { 3 }{ 5 } }={ 3 }^{ 3 }\quad \Rightarrow \left| x \right| =\pm { 3 }^{ 5 }$

$\therefore$ Product of roots of the equation is $\left( { 3 }^{ 5 } \right) \left( { \left( -3 \right) }^{ 5 } \right) =\boxed { { \left( -3 \right) }^{ 10 } }$

Did it exactly the same way, even the variables ;).

pawan dogra - 6 years, 1 month ago

Rajath Naik
Mar 2, 2015

Tried by trial and error method by making a rational guess for x=3^5, so the other real value was evidently x=-3^5..,so product is -3^10..., ans therefore is 13... I do hope there is a proper and systematic soln to this problem..

Is this enough?

Utkarsh Bansal - 6 years, 3 months ago

Come on, you cant just bashed it you have to come up with an answer like Utkarsh.

Jesus Ulises Avelar - 6 years, 3 months ago

Problem 2

For more problems click here

The answer is 13.

2 solutions

0 pending reports