find the zeros of the quadratic polynomial p(x) = 4x2-4x+1
Answers
Step-by-step explanation:
Given :-
p(x) = 4x²-4x+1
To find :-
find the zeros of the quadratic polynomial p(x) ?
Solution :-
Given Polynomial is p(x) = 4x²-4x+1
Splitting the middle term:-
p(x) = 4x²-4x+1
To get the zeroes of p(x) we write it as p(x) = 0
=> 4x²-4x+1 = 0
=> 4x²-2x-2x+1 = 0
=> 2x(2x-1)-1(2x-1) = 0
=> (2x-1)(2x-1) = 0
=> 2x-1 = 0 or 2x-1 = 0
=> 2x = 1
=>x = 1/2
The zeroes are 1/2 and 1/2
Using Identity:-
p(x) = 4x²-4x+1
To get the zeroes of p(x) we write it as p(x) = 0
=> 4x²-4x+1 = 0
=> 2²x² -2×2×x+1² = 0
=> (2x)² -2(x)(2)+1² = 0
This is in the form of a²-2ab+b²
Where, a = 2x and b = 1
We know that
(a-b)² = a²-2ab+b²
We have
(2x-1)² = 0
=> (2x-1)(2x-1) = 0
=> 2x-1 = 0 or 2x-1 = 0
=> 2x = 1
=>x = 1/2
The zeroes are 1/2 and 1/2
Using Sridharacharya formula:-
p(x) = 4x²-4x+1
To get the zeroes of p(x) we write it as p(x) = 0
On Comparing this with the standard quadratic equation ax²+bx+c=0
a = 4
b=-4
c=1
We know that
x = [-b±√(b²-4ac)]/2a
=> x = [-(-4)±√[(-4)²-4(4)(1)]/2(4)
=>x = [4±√(16-16)]/8
=>x = (4±√0)/8
=> x = (4±0)/8
=> x = 4/8
=> x = 1/2
The zeroes are 1/2 and 1/2
Completing the square method :-
p(x) = 4x²-4x+1
To get the zeroes of p(x) we write it as p(x) = 0
=>4x²-4x+1 = 0
On Dividing by 4 both sides then
=> (4x²/4)-(4x/4)+(1/4) = 0/4
=> x²-x+(1/4) = 0
=> x²-x = -1/4
=>x²-(2/2)x=-1/4
=>x²-2(x)(1/2) = -1/4
On adding (1/2)² both sides then
=>x²-2(x)(1/2)+(1/2)² = (-1/4)+(1/2)²
=> [x-(1/2)]² = (-1/4)+(1/4)
Since (a-b)² = a²-2ab+b²
=> [x-(1/2)]² = (-1+1)/4
=> [x-(1/2)]² = 0/4
=>[x-(1/2)]² = 0
=> [x-(1/2)][x-(1/2)] = 0
=> [x-(1/2)]= 0 or [x-(1/2)] = 0
=> x = 1/2
The zeroes are 1/2 and 1/2
Used Methods:-
- Splitting the middle term
- Using Sridharacharya formula
- Completing the square method
- Using Identity
Used formulae:-
- To get the zeroes of p(x) we write it as p(x) = 0
- x = [-b±√(b²-4ac)]/2a
- (a-b)² = a²-2ab+b²
Given :-
4x² - 4x + 1
To Find :-
Zeroes
Solution :-
4x² - 4x + 1
4x² - (2x + 2x) + 1
4x² - 2x - 2x + 1
2(2x - 1) - 1(2x - 1)
(2x - 1) (2x - 1) = 0
2x - 1 = 0 or 2x - 1 = 0
2x = 1
x = 1/2
or
2x = 1
x = 1/2
Zeroes = 1/2 and 1/2
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