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JEE MainMathematicsToughQuadratic Equations

The value of b for which the equation
2log⁡1/25(bx+28)=−log⁡5(12−4x−x2)2\log_{1/25}(bx + 28) = - \log_{5}(12 - 4x - x^{2}) has coincident
real roots if

Show answer and solution
2log⁡5(bx+28)log⁡5(15)2=−log⁡5(12−4x−x2)\frac{2\log_{5}(bx + 28)}{\log_{5}{(\frac{1}{5})}^{2}} = - \log_{5}(12 - 4x - x^{2})
⇒bx+28=12−4x−x2\Rightarrow bx + 28 = 12 - 4x - x^{2}
⇒x2+(b+4)x+16=0\Rightarrow x^{2} + (b + 4)x + 16 = 0
⇒Δ=0\Rightarrow \Delta = 0
(b+4)2=64(b + 4)^{2} = 64
b=4,−12b = 4, - 12

Source: Grass-G / M-16 Mathematics Paper 25 Nov First.docx

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