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Polynomials with New Roots
The quadratic equation 9 t2 – 6 t + 7 = 0 has roots A, B.
A quadratic with roots 2 A – 1, 2 B – 1 is
p2 + p +

tickcross


Worked Solution:

A suitable polynomial is found by the letting p = 2 t – 1 in 9 t2 – 6 t + 7
In this polynomial in t we use the substitution t =
p + 1
---
2
:
9(
p + 1
---
2
)2 =
9
---
4
p2+
9
---
2
p+
9
---
4
–6(
p + 1
---
2
) = –3 p – 3
7 = 7
Total=
9
---
4
p2+
3
---
2
p+
25
---
4
Since any multiple of this polynomial has the same roots, we could multiply it by 4 to give an answer that avoids fractions:
9 p2 + 6 p + 25 but any other multiple is also correct.

© MEI Produced by Dr Ron Knott, 14 August 04 revised 10 April 2007
tom . button [AT] mei . org . uk 		Test reference: PolyNewRoots.html?ref=94894,qu=2