The equation whose roots are greater by 2 than those of x^2 + 3x + 5 = 0
Answers
Step-by-step explanation:
Let the roots of given equation 2x
3
+3x
2
−4x+5=0 is p,q,r
Relation betwen roots and coefficients are
Sum of the roots=
a
−b
∴p+q+r=
a
−b
=
2
−(3)
=
2
−3
Product of the roots=
a
−d
∴p×q×r=
2
−5
Sum of products of the roots taken two at a time=
a
c
∴p×q+p×r+q×r=
a
c
=
2
−4
=−2
Now new roots are p
′
=p+2,q
′
=q+2,r
′
=r+2
New equation is x
3
−(p
′
+q
′
+r
′
)x
2
+((p
′
×q
′
)+(q
′
×r
′
)+(r
′
×p
′
))x−(p
′
×q
′
×r
′
)=0
p
′
+q
′
+r
′
=p+2+q+2+r+2=p+q+r+6=−
2
3
+6=
2
9
p
′
×q
′
×r
′
=(p+2)(q+2)(r+2)=pqr+2(pq+pr+qr)+4(p+q+r)+8=−
2
5
+2(−2)−4×
2
3
+8=−
2
9
(p
′
×q
′
)+(q
′
×r
′
)+(r
′
×p
′
)=(p+2)(q+2)+(r+2)(q+2)+(p+2)(r+2)=pq+qr+pr+12+4p+4q+4r=−2+12+4×
2
−3
=4
So, the equation is x
3
−
2
9
x
2
+4x−(−
2
9
)=0
⇒2x
3
−9x
2
+8x+9=0