Linear Transformation

Algebra Level 4

Let α \alpha and β \beta be the roots of the equation x 2 − 2 x − 2 = 0 x^2-2x-2=0 such that α < β \alpha<\beta .

The value of the α − 2 β \alpha-2\beta can be expressed as − p − q 3 -p-q\sqrt{3} , where p p and q q are positive integers.

Find p + q p+q .


The answer is 4.

Dexter Woo Teng Koon by Dexter Woo Teng Koon , 20, Malaysia

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2 solutions

First we know that α + β = 2 \alpha+\beta=2 and α β = − 2 \alpha\beta=-2

α − 2 β \alpha-2\beta

= − 1 2 ( α + β ) + 3 2 ( α − β ) =-\dfrac{1}{2}(\alpha+\beta)+\dfrac{3}{2}(\alpha-\beta)

= − 1 2 ( 2 ) − 3 2 α 2 + β 2 − 2 α β =-\dfrac{1}{2}(2)-\dfrac{3}{2}\sqrt{\alpha^2+\beta^2-2\alpha\beta}

= − 1 − 3 2 ( α + β ) 2 − 4 α β =-1-\dfrac{3}{2}\sqrt{(\alpha+\beta)^2-4\alpha\beta}

= − 1 − 3 2 4 + 8 =-1-\dfrac{3}{2}\sqrt{4+8}

= − 1 − 3 3 =-1-3\sqrt{3}

∴ p + q = 1 + 3 = 4 \therefore p+q=1+3=4

A α + B β = A + B 2 ( α + β ) + A − B 2 ( α − β ) \boxed { A\alpha+B\beta=\dfrac{A+B}{2}(\alpha+\beta)+\dfrac{A-B}{2}(\alpha-\beta) }

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So that’s where linear transformation is coming from... the line #2...

Jiahui Tan - 1 year, 8 months ago
Sudhamsh Suraj
Mar 7, 2017

Directly from the given equation, ( x − 1 ) 2 (x-1)^2 = 3 .

So roots are 1+√3,1-√3 .

α α =1-√3, β=1+√3. Since α<β .

α - 2β = (1-√3) - 2(1+√3) = -1-3√3.

Therefore, p+q=1+3=4.

1 Helpful 0 Interesting 0 Brilliant 0 Confused

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