Web25 aug. 2015 · now, from above equation, we conclude that q 2 is divisible by 3 which shows that q is also divisible by 3 (by fundamental theorem) then we have (3) q = 3 n Now, from (2) & (3), we find that p & q have a common factor 3 which is a contradiction. Share Cite Follow edited Sep 11, 2015 at 3:30 answered Aug 25, 2015 at 11:25 Harish … Webr3+8r2+22r+20=0 Three solutions were found : r = -2 r = (-6-√-4)/2=-3-i= -3.0000-1.0000i r = (-6+√-4)/2=-3+i= -3.0000+1.0000i Step by step solution : Step 1 :Equation at the end of …

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Web= = pa(a2qr −p2bc)−qb(q2ca− b2pr)+ rc(pqc2 −r2ab) = pqr[a3 +b3 + c3] −abc[p3 + q3 + r3] = pqr[a3 +b3 + c3] −abc[3qpr][∵ p+ q+ r = 0, ∴ p3 + q3 + r3 = 3pqr] = pqr[a3 +b3 + c3 −3abc] = pqr ∣∣a c b b a c c b a∣∣ WebIf p,q,r be three distinct real numbers in A.P. then p 3+r 3 equals A pqr B −3pqr C −6pqr D none of these Medium Solution Verified by Toppr Correct option is D) Given: p,q,r are in A.P. ∴2q=p+r Cubing both sides: ⇒p 3+r 3+3p 2r+3pr 2=8q 3 ∴p 3+r 3=8q 3−3p 2r−3pr 2 Was this answer helpful? 0 0 Similar questions jokes about being fired

If the roots of the quadratic equation p(q - r)x^2 + q(r - p)x + r(p ...

WebIf p, q, r are in AP, then p 3 + r 3 - 8q 3 is equal to -6pqr. Explanation:-∵ p, q, r are in AP. ∴ 2q = p + r. ⇒ p + r – 2q = 0. ∴ p 3 + r 3 + (-2p) 3 = 3 × p × r × -2q [Using if a + 6 + c = 0 … Web16 apr. 2024 · From here, it is clear that if both p ¯ and ( q ∧ r) is false, the complete statement is false. If either is true, then the full statement is true. And so the full … WebIf the roots of the quadratic equation p(q−r)x 2+q(r−p)x+r(p−q)=0 are equal, Show then p1+ r1= q2= qm Find m. Medium Solution Verified by Toppr Correct option is A) p(q−r)x 2+q(r−p)x+r(p−q)=0 D=0∴ the root are equal D=b 2−4ac ⇒(q(r−p)) 2−4(p(q−r))(r(p−q)))=0 ⇒q 2(r 2+p 2−2pr)−4((pq−pr)(pr−qr))=0 ⇒q 2(r 2+p 2−2pr)−4(p 2qr−pq 2r−p 2r 2+pqr 2)=0 jokes about being chatty