This test uses JavaScript to generate your test and mark it. You therefore need to enable Scripting using the Preferences... item in one of the Menu items for this Browser's window. Once enabled, just Reload (Refresh) this page.

Polynomials with New Roots
The quadratic equation t2 – t + 4 = 0 has roots A, B.
A quadratic with roots –2 A + 1, –2 B + 1 is
z2 + z +

tickcross


Worked Solution:

A suitable polynomial is found by the letting z = –2 t + 1 in t2 – t + 4
In this polynomial in t we use the substitution t =
z – 1
---
–2
 =
1 – z
---
2
:
(
1 – z
---
2
)2 =
1
---
4
z2
– 1
---
2
z+
1
---
4
–1(
1 – z
---
2
) =
1
---
2
z
– 1
---
2
4 = 4
Total=
1
---
4
z2+
15
---
4
Since any multiple of this polynomial has the same roots, we could multiply it by 4 to give an answer that avoids fractions:
z2 + 15 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=65467