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Question

If roots of the given quadratic 2x2−9x+7=0 are p,q. Then the equation whose roots are

A
50x245x+7=0
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B
2x243x+203=0
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C
2x2x3=0
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D
2x2+9x+7=0
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Solution

Given: 2x2−9x+7=0; roots are p,q Sum of roots =p+q=92⋯(i) Product of roots =p.q=72⋯(ii) Now we will solve for the one by one a) −p,−q ∴Sum of roots =−p−q=−ba ⇒Sum of roots =−(p+q)=−ba ⇒Sum of roots =−ba=−(92) [From(i)] ⇒Sum of roots =−ba=−92⋯(iii) Now, Product of roots =(−p)(−q)=ca ⇒Productof roots =ca=p.q ⇒Productof roots =ca=72⋯(iv) [From(ii)] Now we have the sum , products of roots for the required equation.And we can write the equation as: x2−(sum of roots)x+(product of roots)=0 ∴x2−(−92)x+(72)=0 ⇒2x2+9x+7=0 b) p−2,q−2 ∴Sum of roots =p−2+q−2=−ba ⇒Sum of roots =(p+q)−4=−ba ⇒Sum of roots =−ba=(92)−4 [From(i)] ⇒Sum of roots =−ba=12⋯(v) Now, Product of roots =(p−2)(q−2)=ca ⇒Productof roots =ca=p.q−2(p+q)+4 ⇒Productof roots =ca=72−2(92)+4 [From(i),(ii)] ⇒Productof roots =ca=−32⋯(vi) Now we have the sum , products of roots for the required equation.And we can write the equation as: x2−(sum of roots)x+(product of roots)=0 ∴x2−(12)x+(−32)=0 ⇒2x2−x−3=0 c) p5,q5 ∴Sum of roots =p5+q5=−ba ⇒Sum of roots =p+q5=−ba ⇒Sum of roots =−ba=((92)5) [From(i)] ⇒Sum of roots =−ba=910⋯(vii) Now, Product of roots =p5.q5=ca ⇒Productof roots =ca=p.q25 ⇒Productof roots =ca=((72)25) [From(ii)] ⇒Productof roots =ca=750⋯(viii) Now we have the sum , products of roots for the required equation.And we can write the equation as: x2−(sum of roots)x+(product of roots)=0 ∴x2−(910)x+(750)=0 ⇒50x2−45x+7=0 d) 3p+4,3q+4 ∴Sum of roots =3p+4+3q+4=−ba ⇒Sum of roots =3(p+q)+8=−ba ⇒Sum of roots =−ba=3(92)+8 [From(i)] ⇒Sum of roots =−ba=432⋯(ix) Now, Product of roots =(3p+4)(3q+4)=ca ⇒Productof roots =ca=9p.q+12(p+q)+16 ⇒Productof roots =ca=9(72)+12(92)+16 [From(ii)] ⇒Productof roots =ca=2032⋯(x) Now we have the sum , products of roots for the required equation.And we can write the equation as: x2−(sum of roots)x+(product of roots)=0 ∴x2−(432)x+(2032)=0 ⇒2x2−43x+203=0

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