find the value of p for which the quadratic equation x[x-4]+p=0 has real roots.
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Hi ,
x ( x - 4 ) + p = 0
x² - 4x + p = 0---( 1 )
compare equation ( 1 ) with ax² + bx + c = 0
a = 1 , b = -4 , c = p
As we know that ,
if the descriminant ≥ 0 then quadratic
equation has real roots .
b² - 4ac ≥ 0
( - 4 )² - 4 × 1 × p ≥ 0
16 - 4p ≥ 0
-4p ≥ - 16
divide both sides with ( - 4 ) we get
( -4p ) / ( - 4 ) ≤ ( - 16 ) / ( -4 )
p ≤ 4
if p ≤ 4 equation ( 1 ) have real roots.
I hope this helps you.
:)
x ( x - 4 ) + p = 0
x² - 4x + p = 0---( 1 )
compare equation ( 1 ) with ax² + bx + c = 0
a = 1 , b = -4 , c = p
As we know that ,
if the descriminant ≥ 0 then quadratic
equation has real roots .
b² - 4ac ≥ 0
( - 4 )² - 4 × 1 × p ≥ 0
16 - 4p ≥ 0
-4p ≥ - 16
divide both sides with ( - 4 ) we get
( -4p ) / ( - 4 ) ≤ ( - 16 ) / ( -4 )
p ≤ 4
if p ≤ 4 equation ( 1 ) have real roots.
I hope this helps you.
:)
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3
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