wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

If roots of the given quadratic 2x29x+7=0 are p,q. Then the equation whose roots are

A
50x245x+7=0
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
2x243x+203=0
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
2x2x3=0
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
2x2+9x+7=0
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

Given: 2x29x+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 =pq=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) p2,q2
Sum of roots =p2+q2=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 =(p2)(q2)=ca

Productof roots =ca=p.q2(p+q)+4

Productof roots =ca=722(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

2x2x3=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

50x245x+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

2x243x+203=0


flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Algebra of Roots of Quadratic Equations
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon