Find value of p so that quadratic equation px(x-3)+9=0 has two equal roots
Answers
Answered by
27
First open the bracket
px^2 -3px+9=0
Condition for equal roots: b^2 - 4ac=0
9p^2- 36p=0
p^2-4p=0
p(p-4)=0
p=0 p=4
px^2 -3px+9=0
Condition for equal roots: b^2 - 4ac=0
9p^2- 36p=0
p^2-4p=0
p(p-4)=0
p=0 p=4
Answered by
12
b²-4ac =o
9p²-4p×9 =0
9p²-36p=0
9p[p-4] =0
p=4
9p²-4p×9 =0
9p²-36p=0
9p[p-4] =0
p=4
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