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