If p,q and r are positive and are in AP, the roots of the quadratic equation px2+qx+r=0 are real for
A
∣∣∣rp−7∣∣∣≥4√3
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B
∣∣pr−7∣∣<4√3
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C
all p and r
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D
no p and r
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Solution
The correct option is A∣∣∣rp−7∣∣∣≥4√3 p,q,r are in AP ∴2q=p+r Since, px2+qx+r=0 has real roots So, D≥0 ∴q2−4pr≥0 ⇒(p+r2)2−4pr≥0 ⇒p2+r2−14pr≥0 ⇒p2r2+1−14pr≥0 ⇒(pr−7)2≥48 ⇒∣∣pr−7∣∣≥4√3