Ppt on mathematician sridharacharya

Solved Examples Using Sridharacharya Formula

Example1.\( 2x^2-2\sqrt{2}x+1=0 \)

Solution: Comparing the given equation with  \(ax^2+bx+c=0 \)

a=2, b= -22, c=1.

Now apply the Sridharacharya formula 

\(\Rightarrow  x={-(-2\sqrt2)\pm\sqrt{(-2\sqrt2)^2-4.2.1}\over{2\times2}} = {2\sqrt2  \pm0\over4}\)

\(\Rightarrow  x={2\sqrt2\pm0\over4} = {2\sqrt2\over4}, {2\sqrt2\over4}={1\over\sqrt2},{1\over\sqrt2}\)

Hence, the given equation has two equal roots: 12 and 12

\({1\over\sqrt2}, {1\over\sqrt2} \)

Example2. 

\({1\over2}x^2-\sqrt{11}x+1=0\)

Solution: Multiplying the given equation by 2 to remove the fraction

\(x^2-2\sqrt{11}x+2=0\)

Now comparing this equation with\( ax^2+bx+c=0\)

a=1, b= -211, c=2.

Now apply the Sridharacharya formula 

\(\Rightarrow x={-(-2\sqrt{11})\pm\sqrt{(-2\sqrt{11})^2-4.2.1}\over{2\times1}} = {2\sqrt{11}\pm\sqrt{36}\over2}\)

x=  22 04 =11+3, 11-3.

\(\Rightarrow  x={2\sqrt2\pm0\over4}=\{/REGREPLACE}

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