Quadratic Equations
Newton's Power Sum — New Quadratic
DAILY_CHALLENGE
Grade 11
Question:
Let $\alpha,\beta$ be the roots of the equation $x^2+2\sqrt{2}x-1=0$. The quadratic equation, whose roots are $\alpha^4+\beta^4$ and $\dfrac{1}{10}(\alpha^6+\beta^6)$, is:
$x^2-190x+9466=0$
$x^2-180x+9506=0$
$x^2-195x+9506=0$
$x^2-195x+9466=0$
Step-by-Step Solution
Key Concept: $\alpha+\beta=-2\sqrt{2}$, $\alpha\beta=-1$. Compute $\alpha^4+\beta^4=(\alpha^2+\beta^2)^2-2(\alpha\beta)^2=((\alpha+\beta)^2-2\alpha\beta)^2-2=(8+2)^2-2=98$. $\alpha^6+\beta^6=970$, so $\frac{1}{10}(\alpha^6+\beta^6)=97$.
$\alpha^4+\beta^4=98$, $\frac{1}{10}(\alpha^6+\beta^6)=97$. $x^2-195x+9506=0$.
Correct Answer: 3