if the quadratic eqution px^_2root5 px + 15 =0 has two equal roots,Find the value of .'p'.
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anju5o:
you both are wrong
answer is 3
plz explain me how you get it plzzz
given, roots are equal. then D=0 , means, b²-4ac=0 : ( -2√5)² - 4(15)(p) =0 : 20p² - 60 p =0 : 20p(p -3)=0 : either , 20p =0 or p-3=0 , then p=0 or 3 . p=0 doesn't follow the equation , therefore p=3
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Hi ,
Compare px² - 2√5px + 15 = 0 with
ax² + bx + c = 0 ,
a = p , b = - 2√5p , c = 15
Determinant = 0
Since it has equal roots .
b² - 4ac = 0
( -2√5p )² - 4 × p × 15 = 0
20p² - 60p = 0
20p( p - 3 ) = 0
20p = 0 or p - 3 = 0
p = 0 or p = 3
I hope this helps you.
:)
Compare px² - 2√5px + 15 = 0 with
ax² + bx + c = 0 ,
a = p , b = - 2√5p , c = 15
Determinant = 0
Since it has equal roots .
b² - 4ac = 0
( -2√5p )² - 4 × p × 15 = 0
20p² - 60p = 0
20p( p - 3 ) = 0
20p = 0 or p - 3 = 0
p = 0 or p = 3
I hope this helps you.
:)
you had done some mistake . correct it
Thank you , anju50
its ok
but your answer is pending
prove that p=3 not 0. ok
I hope you understand
p ≠ 0 because by the definition of quadratic equation
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