Exams Joint Entrance Examination Mathematics Quadratic Equations and Inequalities Question 63 Question 63 of 157 Hard NUMERICAL
Let a , b , c a, b, c a , b , c be the lengths of three sides of a triangle satistying 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 x x is the interval \[(\alpha, \beta)\]
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( a 2 + b 2 ) x 2 − 2 b ( a + c ) x + b 2 + c 2 = 0 ({a}^{2}+{b}^{2}){x}^{2}-2b(a+c)x+{b}^{2}+{c}^{2}=0 ( a 2 + b 2 ) x 2 − 2 b ( a + c ) x + b 2 + c 2 = 0
⇒ a 2 x 2 − 2 a b x + b 2 + b 2 x 2 − 2 b c x + c 2 = 0 ⇒ ( a x − b ) 2 + ( b x − c ) 2 = 0 ⇒ a x − b = 0 , b x − c = 0 ⇒ a + b > c b + c > a c + a > b \begin{gathered} \Rightarrow {a}^{2}{x}^{2}-2abx+{b}^{2}+{b}^{2}{x}^{2}-2bcx+{c}^{2}=0 \\ \Rightarrow (ax-b{)}^{2}+(bx-c{)}^{2}=0 \\ \Rightarrow ax-b=0,\, bx-c=0 \\ \Rightarrow a+b>c\, b+c>a\, c+a>b \end{gathered} ⇒ a 2 x 2 − 2 ab x + b 2 + b 2 x 2 − 2 b c x + c 2 = 0 ⇒ ( a x − b ) 2 + ( b x − c ) 2 = 0 ⇒ a x − b = 0 , b x − c = 0 ⇒ a + b > c b + c > a c + a > b
a + a x > b x a x + b x > a a x 2 + a > a x a + a x > a x 2 a x + a x 2 > a x 2 − x + 1 > 0 x 2 − x − 10 \begin{gathered} a+ax>bx \quad ax+bx>a \quad a{x}^{2}+a>ax \\ a+ax>a{x}^{2} \quad ax+a{x}^{2}>a \quad {x}^{2}-x+1>0 \\ {x}^{2}-x-10 \quad \end{gathered} a + a x > b x a x + b x > a a x 2 + a > a x a + a x > a x 2 a x + a x 2 > a x 2 − x + 1 > 0 x 2 − x − 10
1 − 5 2 − 1 + 5 2 \begin{gathered} \frac{1-\sqrt{5}}{2}\frac{-1+\sqrt{5}}{2} \end{gathered} 2 1 − 5 2 − 1 + 5
⇒ 5 − 1 2 < x < 5 + 1 2 ⇒ α = 5 − 1 2 , β = 5 + 1 2 12 ( α 2 + β 2 ) = 12 ( ( 5 − 1 ) 2 + ( 5 + 1 ) 2 4 ) = 36 \begin{gathered} \Rightarrow \frac{\sqrt{5}-1}{2}<x<\frac{\sqrt{5}+1}{2} \\ \Rightarrow \alpha =\frac{\sqrt{5}-1}{2},\beta =\frac{\sqrt{5}+1}{2} \\ 12({\alpha }^{2}+{\beta }^{2})=12(\frac{(\sqrt{5}-1{)}^{2}+(\sqrt{5}+1{)}^{2}}{4})=36 \end{gathered} ⇒ 2 5 − 1 < x < 2 5 + 1 ⇒ α = 2 5 − 1 , β = 2 5 + 1 12 ( α 2 + β 2 ) = 12 ( 4 ( 5 − 1 ) 2 + ( 5 + 1 ) 2 ) = 36 Answer the question to unlock