# METHOD OF UNDETERMINED COEFFICIENTS

## About "Method of undetermined coefficients"

Method of undetermined coefficients :

Let us look into some example problems on method of undetermined coefficients.

Question 1 :

Factorize: x4 + 1. (Hint: Try completing the square.)

Solution :

x4 + 1  =  (x2√2x + 1)  (x2 - √2x + 1)

Question 2 :

If x2 + x + 1 is a factor of the polynomial 3x3 + 8x2 + 8x + a, then find the value of a.

Solution :

Hence the value of a is 5.

Question 3 :

The equations x2 −6x+a = 0 and x2 −bx+6 = 0 have one root in common. The other root of the first and the second equations are integers in the ratio 4 : 3. Find the common root.

Solution :

Let α be the common root.

Let α, 4β be the roots of x2 − 6x + a = 0.

Let α, 3β be the roots of x2 − bx+6 = 0.

Then, 4αβ = a and 3αβ = 6 which give αβ = 2 and a = 8.

The roots of x2 − 6x+8 = 0 are 2, 4.

If α = 2, then β = 1

If α = 4, then β = 1

2 which is not an integer.

Hence, the common root is 2.

Question 4 :

Find the values of p for which the difference between the roots of the equation x2 + px+8 = 0 is 2.

Solution :

Let α and β be the roots of the equation x2 + px+8 = 0.

Then, α + β = −p, αβ = 8 and |α − β| = 2.

Now, (α + β)2 − 4αβ = (α − β)2, which gives p2 − 32 = 4.

Thus, p = ±6.

After having gone through the stuff given above, we hope that the students would have understood "Method of undetermined coefficients"

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