By factorisation method ( x - …
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Preeti Dabral 2 years, 2 months ago
{tex}Here, we have \begin{aligned} & \Rightarrow(x-3)(x-4)=34 / 33^2 \\ & \Rightarrow x^2-7 x+12-34 / 33^2=0 \\ & \Rightarrow x^2-7 x+13034 / 33^2=0 \\ & \Rightarrow x^2-7 x+98 / 33 \times 133 / 33=0 \\ & \Rightarrow x^2-231 / 33 x+98 / 33 \times 133 / 33=0 \end{aligned} {/tex}
{tex}By using factorization method, we get \begin{aligned} & \Rightarrow x^2-(98 / 33+133 / 33) x+98 / 33 \times 133 / 33=0 \\ & \Rightarrow x^2-98 / 33 x-133 / 33 x+98 / 33 \times 133 / 33=0 \\ & \Rightarrow\left(x^2-98 / 33 x\right)-(133 / 33 x-98 / 33 \times 133 / 33)=0 \\ & \Rightarrow x(x-98 / 33)-133 / 33(x-98 / 33)=0 \\ & \Rightarrow(x-98 / 33)(x-133 / 33)=0 \\ & \Rightarrow x-98 / 33=0 \text { or } x-133 / 33= \\ & \Rightarrow x=98 / 33,133 / 33 \\ & \text { Hence, } x=x=98 / 33,133 / 33 \end{aligned} {/tex}
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