Problem 1 :
Factorize : x4 + 1.
Solution :
x4 + 1 = (x2 + √2x + 1)(x2 - √2x + 1)
Problem 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.
Problem 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.
Problem 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.
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