Question

FG ind the value of p so that the quardatic equation px(x-3)+9=0 has two equal roots

Answer 1

$\sf The \:equation \:is \:px(x-3)+9=0\\\sf </p><p>=px^2-3px+9\\\\\sf </p><p>Now, \:Let \:the\: two\: equal\: root\: of\: the\\\sf quadratic\: equation\: be\: \alpha\\\sf </p><p>Then, \\\sf \longrightarrow</p><p>\alpha+\alpha= \frac{-b} {a} \\\sf \longrightarrow</p><p>2\alpha= \frac{-(-3p)} {p} \\\sf \longrightarrow</p><p>2\alpha= \frac{3\cancel{p}} {\cancel{p}}\\\sf \longrightarrow</p><p>\alpha=\frac{3}{2}\\\sf </p><p>Now, \alpha \:x \:\alpha = \frac{c}{a}\\\sf \longrightarrow</p><p>\alpha^2=\frac{9} {p} \\\sf \longrightarrow</p><p>(\frac{3}{2})^2=\frac{9} {p} \\\sf \longrightarrow</p><p>\frac{9}{4}= \frac{9} {p} \\\sf \longrightarrow</p><p>P=4\\\\</p><p></p><p>\sf Hence, \:p=4,so\: that\: the\:\\ \sf quadratic\: equation\: px(x-3)+9=0 \:has \:two\\ \sf equal\: roots.</p><p>$