Find the values of p for which the equation 3x2 – px + 5 = 0 has real roots.
Answers
Answered by
54
3x² - px + 5 = 0 has real roots
So p² - 4*3*5 ≥ 0
⇒ p² ≥ 60
⇒ p ≤ -√60 and p ≥ √60
So p² - 4*3*5 ≥ 0
⇒ p² ≥ 60
⇒ p ≤ -√60 and p ≥ √60
kabir35:
could u explain the last step.
p^2 > 60 .....Thus |p| will be greater than √60
if |p| > √60, then p is either greater than √60 or less than -√60
mathematically, p<-√60 and p>√60
p belongs to (-infinity, -√60] U [√60, infinity)
Answered by
46
Given Equation is 3x^2 - px + 5 = 0
a = 3,b = -p,c = 5.
Given that the equation has real roots.
We know that if the roots are equal then the discriminant is 0.
b^2 - 4ac = 0
(-p)^2 - 4(3)(5) = 0
-p^2 - 60 = 0
p^2 = 60
p = 2root 15 (or) -2root 15.
Hope this helps!
a = 3,b = -p,c = 5.
Given that the equation has real roots.
We know that if the roots are equal then the discriminant is 0.
b^2 - 4ac = 0
(-p)^2 - 4(3)(5) = 0
-p^2 - 60 = 0
p^2 = 60
p = 2root 15 (or) -2root 15.
Hope this helps!
p.s. the question tells only real roots not equal!!!
The formula is same
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