for what value of'p' will the quadratic equation have real roots? px2+4x+1=0
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Answered by
120
Hi ,
compare px² + 4x + 1 = 0 with ax² + bx + c = 0,
a = p , b = 4 , c = 1
we know that if the discreaminant ≥ 0 then
the quadratic equation has real roots.
b² - 4ac ≥ 0
4² - 4 × p × 1 ≥ 0
16 - 4p ≥ 0
- 4p ≥ - 16
multiply both sides with ( -1 ) , we get
4p ≤ 16
p ≤ 16 / 4
p ≤ 4
therefore ,
when p ≤ 4 then the quadratic equation have
real roots .
I hope this helps you.
:)
compare px² + 4x + 1 = 0 with ax² + bx + c = 0,
a = p , b = 4 , c = 1
we know that if the discreaminant ≥ 0 then
the quadratic equation has real roots.
b² - 4ac ≥ 0
4² - 4 × p × 1 ≥ 0
16 - 4p ≥ 0
- 4p ≥ - 16
multiply both sides with ( -1 ) , we get
4p ≤ 16
p ≤ 16 / 4
p ≤ 4
therefore ,
when p ≤ 4 then the quadratic equation have
real roots .
I hope this helps you.
:)
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27
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