Quadratic Equations
Triangle Inequality — Range of x and Sum of Squares
nta_pyq_2024_jan
Grade 11
Question:
Let $a,b,c$ be the lengths of three sides of a triangle satisfying the condition $(a^2+b^2)x^2-2b(a+c)x+(b^2+c^2)=0$. If the set of all possible values of $x$ is the interval $(\alpha,\beta)$, then $12(\alpha^2+\beta^2)$ is equal to
Step-by-Step Solution
Key Concept: $(a^2+b^2)x^2-2b(a+c)x+(b^2+c^2)=0\Rightarrow(ax-b)^2+(bx-c)^2=0\Rightarrow ax=b$ and $bx=c\Rightarrow x=b/a=c/b$. For triangle inequality constraints, find the range of $x=b/a$.
$\alpha=\frac{\sqrt5-1}{2}$, $\beta=\frac{\sqrt5+1}{2}$. $12(\alpha^2+\beta^2)=12\times3=36$.
Correct Answer: 36