Quadratic Equation Question 10
Question 10 - 2024 (31 Jan Shift 2)
Let $a, b, c$ be the length of three sides of a triangle satisfying the condition $\left(a^{2}+b^{2}\right) x^{2}-2 b(a+c)$.
$x+\left(b^{2}+c^{2}\right)=0$. If the set of all possible values of $x$ is the interval $(\alpha, \beta)$, then $12\left(\alpha^{2}+\beta^{2}\right)$ is equal to
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Answer (36)
Solution
$$ \left(a^{2}+b^{2}\right) x^{2}-2 b(a+c) x+b^{2}+c^{2}=0 $$
$\Rightarrow a^{2} x^{2}-2 a b x+b^{2}+b^{2} x^{2}-2 b c x+c^{2}=0$
$\Rightarrow(a x-b)^{2}+(b x-c)^{2}=0$
$\Rightarrow a x-b=0, \quad b x-c=0$
$\Rightarrow a+b>c \quad b+c>a \quad c+a>b$
$a+a x>b x \quad|a x+b x>a| a x^{2}+a>a x$
$a+a x>a x^{2} \left\lvert, \begin{aligned} & a x+a x^{2}>a \ & x^{2}-x+1>0\end{aligned}\right.$
$x^{2}-x-1<0\left|x^{2}+x-1>0\right|$ always true
$\frac{1-\sqrt{5}}{2}<x<\frac{1+\sqrt{5}}{2}$
$x<\frac{-1-\sqrt{5}}{2}$, or $x>\frac{-1+\sqrt{5}}{2}$
$\Rightarrow \frac{\sqrt{5}-1}{2}<\mathrm{x}<\frac{\sqrt{5}+1}{2}$
$\Rightarrow \alpha=\frac{\sqrt{5}-1}{2}, \beta=\frac{\sqrt{5}+1}{2}$
$2\left(\alpha^{2}+\beta^{2}\right)=12\left(\frac{(\sqrt{5}-1)^{2}+(\sqrt{5}+1)^{2}}{4}\right)=36$
