જો સમીકરણો ax^2 + bx + c = 0 અને px^2 + qx + | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

${\alpha _1}\, + \,{\alpha _2}\, = \,\,\frac{{ - b}}{a},\,\,{\alpha _1}{\alpha _2}\, = \,\,\frac{c}{a}$

${\beta _1}\, + \,\,{\beta _2}\,\, = \,\,\f

Vedclass · Accepted Answer

$b^2pr = q^2ac$

Vedclass · Answer

${\alpha _1}\, + \,{\alpha _2}\, = \,\,\frac{{ - b}}{a},\,\,{\alpha _1}{\alpha _2}\, = \,\,\frac{c}{a}$

${\beta _1}\, + \,\,{\beta _2}\,\, = \,\,\frac{{ - q}}{p},\,\,{\beta _1}{\beta _2}\, = \,\,\frac{r}{p}$

જ્યાં સમીકરણો ${\alpha _{m{1}}}y\,\, + \,\,{\alpha _2}z\,\, = \,\,0$ અને 

$\,{\beta _1}y\,\, + \,\,{\beta _2}z\,\, = \,\,0$ શૂન્યેતર ઉકેલ ધરાવે છે.

$	herefore \,\left| {\begin{array}{*{20}{c}}
{{\alpha _1}}&{{\alpha _2}}\
{{\beta _1}}&{{\beta _2}}
\end{array}} ight|\, = \,\,0\,\,$

$ \Rightarrow \,{\alpha _1}{\beta _2}\,\, - \,\,\,{\alpha _2}{\beta _1}\, = \,0$

$\frac{{{\alpha _1}}}{{{\beta _1}}}\,\, = \,\,\frac{{{\alpha _2}}}{{{\beta _2}}}\,\, \Rightarrow \,\,\frac{{{\alpha _1}\, + \,{\alpha _2}}}{{{\beta _1} + {\beta _2}}}\, = \,\, \mp \,\sqrt {\left\{ {\frac{{{\alpha _1}\,{\alpha _2}}}{{{\beta _1}{\beta _2}}}} ight\}} $

$	herefore \,\,\,\frac{{\frac{{ - b}}{a}}}{{\frac{{ - q}}{p}}}\, = \,\sqrt {\left\{ {\frac{{\frac{c}{a}}}{{\frac{r}{p}}}} ight\}} \,$

$ \Rightarrow \,\frac{{{b^2}\,{p^2}}}{{{q^2}{a^2}}}\,\, = \,\,\frac{{cp}}{{ar}}\,$

$ \Rightarrow \,{b^2}pr\,\, = \,\,{q^2}\,ac$

Similar Questions

સમીકરણ સંહિતા $x+y+z=\beta $ , $5x-y+\alpha z=10$ , $2x+3y-z=6$ ના અનન્ય ઉકેલ ......... પર આધારિત છે

જો રેખીય સમીકરણો $x + y + z = 5$ ; $x = 2y + 2z = 6$ ; $x + 3y + \lambda z = u (\lambda \, \mu \in R)$ અનંત ઉકેલ ધરાવે છે તો $\lambda + \mu $ ની કિમંત મેળવો.

ધારો કે સમીકરણ સંહતિ $x+y+k z=2$ ; $2 x+3 y-z=1$ ; $3 x+4 y+2 z=k$ ને અસંખ્ય ઉકેલો છે. $( k +1) x +(2 k -1) y =7$ ; $(2 k +1) x +( k +5) y =10$ ને:

જો $\left| {\,\begin{array}{*{20}{c}}a&b&{a + b}\\b&c&{b + c}\\{a + b}&{b + c}&0\end{array}\,} \right| = 0$; તો $a,b,c$ એ .. . . શ્રેણીમાં છે .

સમીકરણ $\left| {\,\begin{array}{*{20}{c}}x&2&{ - 1}\\2&5&x\\{ - 1}&2&x\end{array}\,} \right| = 0$ નો ઉકેલ મેળવો.