Question

the area of a triangle formed by (0,0) , (a^x^2,0) , (0,a^6x) is 1/2a^5 sq.units then x=

Answer 1

$\textbf{Given:}$

$\text{Area of the triangle formed by (0,0),(a^{x^2},0),(0,a^{6x}) is \dfrac{1}{2}a^5 sq.units}$

$\textbf{To find:}$

$\text{The value of x}$

$\textbf{Solution:}$

$\text{Since area of the triangle formed by (0,0),(a^{x^2},0),(0,a^{6x}) is 1/2a^5 sq.units, we have}$

$\bf\dfrac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]=\dfrac{1}{2}a^5$

$\dfrac{1}{2}[0(0-a^{6x})+a^{x^2}(a^{6x}-0)+0(0-0)]=\dfrac{1}{2}a^5$

$\dfrac{1}{2}[a^{x^2}(a^{6x})]=\dfrac{1}{2}a^5$

$a^{x^2}(a^{6x})=a^5$

$a^{x^2+6x}=a^5$

$\text{Equating powers on bothsides, we get}$

$x^2+6x=5$

$x^2+6x-5=0$

$\implies\,x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

$\implies\,x=\dfrac{-6\pm\sqrt{36+20}}{2}$

$\implies\,x=\dfrac{-6\pm\sqrt{56}}{2}$

$\implies\,x=\dfrac{-6\pm2\sqrt{14}}{2}$

$\implies\,x=-3\pm\sqrt{14}$

$\therefore\textbf{The values of x are}$

$\bf\,-3+\sqrt{14},\,-3-\sqrt{14}$

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