JEE Main 2025 23 January Morning Shift
Exam JEE Main
Year 2025
Date 23 January
Shift Morning Shift
Subject Mathematics
Chapter Quadratic Equation JEE Main 2025 · 23 January · Morning Shift · Mathematics · Quadratic Equation
If the equation a(b-c) x^2+b(c-a) x+c(a-b)=0 has equal roots, where...
If the equation
a ( b − c ) x 2 + b ( c − a ) x + c ( a − b ) = 0 \mathrm{a}(\mathrm{b}-\mathrm{c}) \mathrm{x}^2+\mathrm{b}(\mathrm{c}-\mathrm{a}) \mathrm{x}+\mathrm{c}(\mathrm{a}-\mathrm{b})=0 a ( b − c ) x 2 + b ( c − a ) x + c ( a − b ) = 0 has equal roots, where
a + c = 15 \mathrm{a}+\mathrm{c}=15 a + c = 15 and
b = 36 5 \mathrm{b}=\frac{36}{5} b = 5 36 , then
a 2 + c 2 a^2+c^2 a 2 + c 2 is equal to
Step-by-step solution
a ( b − c ) x 2 + b ( c − a ) x + c ( a − b ) = 0 x = 1 is root ∴ other root is 1 α + β = − b ( c − a ) a ( b − c ) = 2 ⇒ − b c + a b = 2 a b − 2 a c ⇒ 2 a c = a b + b c ⇒ 2 a c = b ( a + c ) ⇒ 2 a c = 15 b … ( 1 ) ⇒ 2 a c = 15 ( 36 5 ) = 108 ⇒ a c = 54 a + c = 15 a 2 + c 2 + 2 a c = 225 a 2 + c 2 = 225 − 108 = 117 \begin{aligned} & a(b-c) x^2+b(c-a) x+c(a-b)=0 \\ & x=1 \text{ is root } \therefore \text{ other root is } 1 \\ & \alpha+\beta=-\frac{b(c-a)}{a(b-c)}=2 \\ & \Rightarrow-\mathrm{bc}+\mathrm{ab}=2 \mathrm{ab}-2 \mathrm{ac} \\ & \Rightarrow 2 \mathrm{ac}=\mathrm{ab}+\mathrm{bc} \\ & \Rightarrow 2 \mathrm{ac}=\mathrm{b}(\mathrm{a}+\mathrm{c}) \\ & \Rightarrow 2 \mathrm{ac}=15 \mathrm{ b} \ldots(1) \\ & \Rightarrow 2 \mathrm{ac}=15\left(\frac{36}{5}\right)=108 \\ & \Rightarrow \mathrm{ac}=54 \\ & \mathrm{a}+\mathrm{c}=15 \\ & \mathrm{a}^2+\mathrm{c}^2+2 \mathrm{ac}=225 \\ & \mathrm{a}^2+\mathrm{c}^2=225-108=117\end{aligned} a ( b − c ) x 2 + b ( c − a ) x + c ( a − b ) = 0 x = 1 is root ∴ other root is 1 α + β = − a ( b − c ) b ( c − a ) = 2 ⇒ − bc + ab = 2 ab − 2 ac ⇒ 2 ac = ab + bc ⇒ 2 ac = b ( a + c ) ⇒ 2 ac = 15 b … ( 1 ) ⇒ 2 ac = 15 ( 5 36 ) = 108 ⇒ ac = 54 a + c = 15 a 2 + c 2 + 2 ac = 225 a 2 + c 2 = 225 − 108 = 117
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