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Polynomials with New Roots
The quadratic equation 4 t2 – 8 t – 3 = 0 has roots A, B.
A quadratic with roots –2 A + 3, –2 B + 3 is
y2 + y +

tickcross


Worked Solution:

A suitable polynomial is found by the letting y = –2 t + 3 in 4 t2 – 8 t – 3
In this polynomial in t we use the substitution t =
y – 3
---
–2
 =
3 – y
---
2
:
4(
3 – y
---
2
)2 = y2 – 6 y + 9
–8(
3 – y
---
2
) = 4 y – 12
–3 = –3
Total=y2 – 2 y – 6
y2 – 2 y – 6 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=35304,qu=23