Quadratic Equation (Discriminant)

 (CIE 0606/2021/m/22/Q2)
Find the values of the constant $k$ for which the equation $k x^{2}-3(k+1) x+25=0$ has equal roots.

$[4]$

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*********math solution*************

$\begin{array}{ll}&b^{2}-4 a c=0, \text { where } a=k, b=-3(k+1), c=25 \\&\therefore[-3(k+1)]^{2}-4 k(25)=0 \\&9\left(k^{2}+2 k+1\right)-100 k=0 \\&9 k^{2}+18 k+9-100 k=0 \\&9 k^{2}-82 k+9=0 \\&(q k-1)(k-q)=0 \\&9 k-1=0, \quad \text { or } \quad k-9=0 \\&k=\dfrac{1}{9} \quad \text { or } \quad k=9\end{array}$
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