If p-q are the roots of the quadratic equation x 2 -2bx-c-0- prove that 2log - y-p - y-q -log 2 -log y-b- y 2 -2by-c

p+q= 2b;pq=c 2log(√(y p)+√(y q)) =log(2y p q+2√((y p)(y q))) =log2(y+b+√(y^2+2by+c)) =log 2+log(y+b+√(y^2+2by+c)) ...

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Standard XII

Mathematics

Domain and Range of Trigonometric Ratios

If p,q are ...

Question

If $p, q$ are the roots of the quadratic equation $x^{2} + 2 b x + c = 0$ , prove that $2 log (\sqrt{y - p} + \sqrt{y - q}) = log 2 + log (y + b + \sqrt{y^{2} + 2 b y + c})$

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Solution

$p + q = - 2 b; p q = c$
$2 log (\sqrt{y - p} + \sqrt{y - q}) = log (2 y - p - q + 2 \sqrt{(y - p) (y - q)}) = log 2 (y + b + \sqrt{y^{2} + 2 b y + c}) = log 2 + log (y + b + \sqrt{y^{2} + 2 b y + c})$

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If p,q are the roots of the quadratic equation x2+2bx+c=0, prove that 2log(√y−p+√y−q)=log2+log(y+b+√y2+2by+c)

p+q=−2b;pq=c2log(√y−p+√y−q)=log(2y−p−q+2√(y−p)(y−q))=log2(y+b+√y2+2by+c)=log2+log(y+b+√y2+2by+c)

If $p, q$ are the roots of the quadratic equation $x^{2} + 2 b x + c = 0$ , prove that $2 log (\sqrt{y - p} + \sqrt{y - q}) = log 2 + log (y + b + \sqrt{y^{2} + 2 b y + c})$

$p + q = - 2 b; p q = c$
$2 log (\sqrt{y - p} + \sqrt{y - q}) = log (2 y - p - q + 2 \sqrt{(y - p) (y - q)}) = log 2 (y + b + \sqrt{y^{2} + 2 b y + c}) = log 2 + log (y + b + \sqrt{y^{2} + 2 b y + c})$