If the roots of the equation x2-k x-m-0 is tan 2 x and 2 x- then the minimum possible value of k-m is

Roots are tan^2x and ^2x We know that nbsp;tan^2x+^2x= ba=k Also, nbsp;tan^2x·^2x=ca=m ⇒tan^2x·^2x=1=m tan^2x+1tan^2x=k Let tan^2x =X ⇒ X+1X=k We know that ...

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Standard XIII

Mathematics

Relation between Roots and Coefficients for Quadratic

If the roots ...

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If the roots of the equation $x^{2} - k x + m = 0$ is ${tan}^{2} x$ and ${cot}^{2} x,$ then the minimum possible value of $k + m$ is

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Solution

Roots are ${tan}^{2} x$ and ${cot}^{2} x$
We know that ${tan}^{2} x + {cot}^{2} x = \frac{- b}{a} = k$
Also, ${tan}^{2} x \cdot {cot}^{2} x = \frac{c}{a} = m$
$\Rightarrow {tan}^{2} x \cdot {cot}^{2} x = 1 = m$

${tan}^{2} x + \frac{1}{{tan}^{2} x} = k$
Let ${tan}^{2} x = X$
$\Rightarrow X + \frac{1}{X} = k$
We know that the minimum value of $X + \frac{1}{X}$ is $2$ , where $X > 0$
$∴$ Minimum value of $k + m$ is $2 + 1 = 3$

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If the roots of the equation x2−kx+m=0 is tan2x and cot2x, then the minimum possible value of k+m is

Roots are tan2x and cot2x We know that tan2x+cot2x=−ba=k Also, tan2x⋅cot2x=ca=m ⇒tan2x⋅cot2x=1=m tan2x+1tan2x=k Let tan2x=X ⇒X+1X=k We know that the minimum value of X+1X is 2, where X>0 ∴ Minimum value of k+m is 2+1=3

If the roots of the equation $x^{2} - k x + m = 0$ is ${tan}^{2} x$ and ${cot}^{2} x,$ then the minimum possible value of $k + m$ is