Question 1 :
Find the zeros of the polynomial function f(x) = 4x2 − 25.
Question 2 :
If x = −2 is one root of x3 − x2 − 17x = 22, then find the other roots of equation.
Question 3 :
Find the real roots of x4 = 16.
Question 4 :
Solve (2x + 1)2 − (3x + 2)2 = 0
Question 1 :
Find the zeros of the polynomial function f(x) = 4x2 − 25.
Solution :
f(x) = 0
4x2 − 25 = 0
4x2 = 25
x2 = 25/4
x = √(25/4)
x = ±5/2
x = 5/2 or -5/2
Question 2 :
If x = −2 is one root of x3 − x2 − 17x = 22, then find the other roots of equation.
Solution :
Given that : x3 − x2 − 17x = 22
x3 − x2 − 17x - 22 = 0
One root of the given polynomial is x = -2
From this, we may find the factor (x + 2)
By using synthetic division, we may find other two factors.
So, the factors are (x + 2) (x2 − 3x - 11)
By solving x2 − 3x - 11, we may get other two values.
x = [-b ± √(b2 - 4ac)] / 2a
a = 1, b = -3 and c = -11
x = [3 ± √(32 - 4(1)(-11))] / 2(1)
x = [3 ± √(9 + 44)] / 2
x = [3 ± √53] / 2
Hence the other roots are (3 + √53)/2 and (3 + √53)/2.
Question 3 :
Find the real roots of x4 = 16.
Solution :
x4 = 16
x4 = 24
x = ±2
Question 4 :
Solve (2x + 1)2 − (3x + 2)2 = 0
Solution :
(2x + 1)2 − (3x + 2)2 = 0
Expanding the above expression using algebraic identities, we get
(2x + 1) + (3x + 2) = 0 (or) (2x + 1) - (3x + 2) = 0
2x + 3x + 1 + 2 = 0 5x + 3 = 0 x = -3/5 |
2x - 3x + 1 - 2 = 0 -x - 1 = 0 -x = 1 x = -1 |
Hence the solutions are -3/5 and -1.
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