Math, asked by Sree8328, 3 months ago

29.Find the value of p ,for which the quadratic equation px2+6x+1 =0 has real roots?

Answers

Answered by jayeshmaya123
0

Answer:

Consider x

2

+px−1=0

Discriminant D=p

2

−4(1)(−1)=p

2

+4

We know p

2

≥0 for all values of p

⇒p

2

+4≥0 (since 4>0)

Therefore D≥0

Hence the equation x

2

+px−1=0 has real and distinct roots for all real values of p.

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