find the value of p so that the quadratic equation px(x-3)+9=0 has two real roots.
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Answered by
3
★ QUADRATIC RESOLUTION ★
GIVEN EQUATION : px ( x - 3 ) + 9 = 0
px² - 3px + 9 = 0
For obtaining two real roots , D ≥ 0
Resolving for equality :
D = 0
9p² - 4 ( p ) 9 = 0
p² - 4p = 0
p ( p - 4 ) = 0
Either p = 0 , 4
★✩★✩★✩★✩★✩★✩★✩★✩★✩★✩★
GIVEN EQUATION : px ( x - 3 ) + 9 = 0
px² - 3px + 9 = 0
For obtaining two real roots , D ≥ 0
Resolving for equality :
D = 0
9p² - 4 ( p ) 9 = 0
p² - 4p = 0
p ( p - 4 ) = 0
Either p = 0 , 4
★✩★✩★✩★✩★✩★✩★✩★✩★✩★✩★
Answered by
0
p=0 or p=4
Step-by-step explanation:
px(x-3) + 9 = 0
px2-3px+9=0
a=p , b=-3p , c=9
b2-4ac= (-3p)2-4(p)(9)
=9p2-36p
for having equal roots
b2-4ac = 9p2-36p = 0
9p2-36p = 0
9p(p-4) = 0
9p=0 (or) p-4=0
p=0(rejected) or p=4
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