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