# ALGEBRA PUZZLES - SOLVING POLYNOMIAL EQUATIONS OF HIGHER DEGREE WITH EXAMPLES

Method of Solving Equations before Algebra Puzzles
if you have not already done so.
There we gave introduction to Algebra
Equations of higher degree and discussed
the method of solving them.

We developed the relations between
the roots and coefficients of the
equation and gave Formulas for the same.
That knowledge is a prerequisite here.
So please study them before proceeding further.

Now Problems on Algebra Puzzles follow.

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## Example 1 : Algebra Puzzles

Solve the following problem on Algebra Puzzles.

If the Algebra Equation x4 + 4x3 - 2x2 - 12x + 9 = 0, has a pair of equal roots, solve the equation.

Solution to Example 1 on Algebra Puzzles :
The given Algebra Equation is
x4 + 4x3 - 2x2 - 12x + 9 = 0
Comparing this with x4 + p1x3 + p2x2 + p3x + p4 = 0, we get
p1 = 4; p2 = -2; p3 = -12; p4 = 9

Also we have
s1 = -p1 = -4;s2 = p2 = -2;s3 = -p3 = 12;s4 = p4 = 9;

By data two pairs of roots are equal.
∴ Let the roots be α, α, β, β
s1 = α + α + β + β = 2(α + β) = -4⇒ α + β = -2........(i)

s4 = (α)(α)(β)(β) = (αβ)2 = 9⇒ αβ = ±3 ......(ii)

Using (i) and (ii), we get
(α - β)2 = (α + β)2 - 4αβ= (-2)2 - 4(±3) = 4 - 12 or 4 + 12 = -8 or 16
Taking positive value, α - β = ±4.......(iii)
(i) + (iii) gives 2α = 2 or -6⇒ α = 1 or -3
using these in (i), we get
β = -2 - α = -2 -1 or -2 +3 = -3 or 1
These are the same as α values.
∴ the roots are 1, 1, -3, -3.

These roots are found using s1 and s4.
Let us verify whether these satisfy s2 and s3.
s2 = (1)(1) + (1)(-3) + (1)(-3) + (1)(-3) + (1)(-3) + (-3)(-3)
= 10 - 12 = -2 [satisfied.]
s3 = (1)(1)(-3) + (1)(1)(-3) + (1)(-3)(-3) + (1)(-3)(-3) = 18 - 6 = 12 [satisfied.]

Thus the roots of the given Algebra Equation are
1, 1, -3, -3. Ans.

Thus Example 1 on Algebra Puzzles is solved.

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## Example 2 : Algebra Puzzles

Solve the following problem on Algebra Puzzles.

Solve the Algebra Equation 16x4 - 64x3 + 56x2 + 16x - 15 = 0, the roots of which are in A.P.

Solution to Example 2 on Algebra Puzzles :
The given Algebra Equation is 16x4 - 64x3 + 56x2 + 16x - 15 = 0
Dividing both sides of the equation by 16, we get
x4 - 4x3 + (7⁄2)x2 + x - 15⁄16 = 0
Comparing this with x4 + p1x3 + p2x2 + p3x + p4 = 0, we get
p1 = -4; p2 = 7⁄2; p3 = 1; p4 = -15⁄16

By data the roots are in A.P.
Let the roots be α - 3β, α - β, α + β, α + 3β
We know, s1 = -p1 = 4⇒ α - 3β + α - β + α + β + α + 3β = 4⇒ 4α = 4⇒ α = 1......(i)

Using the value of α = 1, the roots become 1 - 3β, 1 - β, 1 + β, 1 + 3β

We know, s2 = p2 = 7⁄2⇒ (1 - 3β)(1 - β) + (1 - 3β)(1 + β) + (1 - 3β)(1 + 3β) + (1 - β)(1 + β) + (1 - β)(1 + 3β) + (1 + β)(1 + 3β) = 7⁄2
⇒ (1 - 4β + 3β2) + (1 - 2β - 3β2) + (1 - 9β2) + (1 - β2) + (1 + 2β - 3β2) + (1 + 4β + 3β2) = 7⁄2
⇒ 6 - 10β2 = 7⁄2 ⇒ 12 - 20β2 = 7
⇒ 20β2 - 5 = 0 ⇒ β2 = 1⁄4⇒ β = ±(1⁄2)......(ii)

When β = +(1⁄2), the roots become -1⁄2, 1⁄2, 3⁄2, 5⁄2.
When β = -(1⁄2), the roots become 5⁄2, 3⁄2, 1⁄2, -1⁄2.
As we can see, both sets are same.

Let us verify whether the set of roots satisfy values of s3 and s4We know, s3 = -p3 = -1 and s4 = p4 = -15⁄16But, s3 = (-1⁄2)(1⁄2)(3⁄2) + (-1⁄2)(1⁄2)(5⁄2) + (-1⁄2)(3⁄2)(5⁄2) + (1⁄2)(3⁄2)(5⁄2)
= (-3⁄8) + (-5⁄8) + (-15⁄8) + (15⁄8) = -1 [satisfied.]s4 = (-1⁄2)(1⁄2)(3⁄2)(5⁄2) = -15⁄16 [satisfied.]

∴ The roots of the given Algebra Equation are
-1⁄2, 1⁄2, 3⁄2, 5⁄2. Ans.

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## Exercise : Algebra Puzzles

Solve the following problems on Algebra Puzzles.

1. Solve the Algebra Equation 9x3 - 15x2 + 7x - 1 = 0 two of the roots being equal..
2. Solve the Algebra Equation x4 - 2x3 - 21x2 + 22x + 40 = 0, the roots of which are in A.P.

For Answers see at the bottom of the page.

## Answers to Exercise : Algebra Puzzles

Answers to problems on Algebra Puzzles.

1. 1⁄3, 1⁄3, 1
2. -4, -1, 2, 5

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