Quadradica!

Algebra Level 3

$\large f(x) = \alpha x^{2} - (\alpha^{2} + 1)x + \alpha$

If function $f(x)$ is defined as above, where $\alpha$ is real , find the roots of $f(y)$ .

\alpha^{2}, -y^{2}

1, 0

y,-y

\alpha, \frac{1}{\alpha}

\pm \sqrt [ 2 ]{ \alpha }

2, 4

1 solution

Viki Zeta
Jun 18, 2016

$f(x)\quad =\quad { \alpha x }^{ 2 }\quad -\quad ({ a }^{ 2 }\quad +\quad 1)x\quad -\quad \alpha \\ f(x)\quad =\quad \alpha { x }^{ 2 }\quad -\quad { \alpha }^{ 2 }x\quad +\quad x\quad -\quad a\\ \quad \quad \quad \quad =\quad \alpha x({ x }\quad -\quad \alpha )\quad +\quad 1(x\quad -\quad \alpha )\\ \qquad \quad =\quad (\alpha x\quad +\quad 1)(x\quad -\quad a)\\ \\ So,\quad The\quad root\quad of\quad above\quad is;\\ \alpha x\quad +\quad 1\quad =\quad 0\\ x\quad =\quad \frac { 1 }{ \alpha } \\ x\quad -\quad \alpha \quad =\quad 0\\ x\quad =\quad \alpha \\$

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