Find the zeroes of the following quadratic polynomials and verify the basic relationship between the zeroes and coefficients.
(i) x2–2x–8 (ii) 4x2–4x+1
(iii) 6x2–7x-3 (iv) 4x2+8x
(v) x2 - 15
(i) Answer :
Let p(x) = x2–2x–8
To find zeroes, we can equate p(x) to 0.
So, p(x) = 0
By factoring, we get
(x-4) (x+2) = 0
Verifying whether they are zeroes :
x-4 = 0 x = 4 p(4) = (4-4) (4+2) p(4) = 0 |
x+2 = 0 x = -2 p(-2) = (-2-4) (-2+2) p(-2) = 0 |
So, 4 and -2 are zeroes of the given quadratic polynomial.
So, α = 4 and β = -2
By comparing the quadratic equation with
ax2+bx+c = 0
a = 1, b = 2 and c = -8
Verifying the relationship :
The sum of the zeroes (α+β) :
4-2 = -b/a
2 = -(-2)/1
2 = 2
The product of the zeroes αβ :
4(-2) = c/a
-8 = -8/1
-8 = -8
Thus, the basic relationship verified.
(ii) Answer :
Let p(x) = 4x2–4x+1
To find zeroes, we can equate p(x) to 0.
So, p(x) = 0
By factoring, we get
(2x-1) (2x-1) = 0
Verifying whether they are zeroes :
2x-1 = 0
x = 1/2
p(1/2) = (2(1/2)-1) (2(1/2)-1)
p(1/2) = 0
So, 1/2 is zero of the given quadratic polynomial.
So, α = 1/2 and β = 1/2
By comparing the quadratic equation with
ax2+bx+c = 0
a = 4, b = -4 and c = 1
Verifying the relationship :
The sum of the zeroes (α+β) :
1/2 + 1/2 = 4/4
1 = 1
The product of the zeroes αβ :
1/2(1/2) = 1/4
1/4 = 1/4
Thus, the basic relationship verified.
(iii) Answer :
Let p(x) = 6x2–7x-3
To find zeroes, we can equate p(x) to 0.
So, p(x) = 0
By factoring, we get
(2x-3) (3x+1) = 0
Verifying whether they are zeroes :
2x-3 = 0 x = 3/2 p(3/2) = (2x-3) (3x+1) p(3/2) = (2(3/2)-3) (3(3/2)+1) p(3/2) = 0 |
3x+1 = 0 x = -1/3 p(-1/3) = (2(-1/3)-3) (3(-1/3)+1) p(-1/3) = 0 |
So, 3/2 and -1/3 are zeroes of the given quadratic polynomial.
So, α = 3/2 and β = -1/3
By comparing the quadratic equation with
ax2+bx+c = 0
a = 6, b = -7 and c = -3
Verifying the relationship :
The sum of the zeroes (α+β) :
3/2 - 1/3 = 7/6
(9-2)/6 = 7/6
7/6 = 7/6
The product of the zeroes αβ :
(3/2)(-1/3) = -3/6
-1/2 = -1/2
Thus, the basic relationship verified.
(iv) Answer :
Let p(x) = 4x2+8x
To find zeroes, we can equate p(x) to 0.
So, p(x) = 0
By factoring, we get
4x(x+2) = 0
Verifying whether they are zeroes :
4x = 0 x = 0 p(0) = 4(0) (0+2) p(0) = 0 |
x+2 = 0 x = -2 p(-2) = 4(-2)(-2+2) p(0) = 0 |
So, 0 and -2 are zeroes of the given quadratic polynomial.
So, α = 0 and β = -2
By comparing the quadratic equation with
ax2+bx+c = 0
a = 4, b = 8 and c = 0
Verifying the relationship :
The sum of the zeroes (α+β) :
0-2 = -8/4
-2 = -2
The product of the zeroes αβ :
(0)(-2) = 0/4
0 = 0
Thus, the basic relationship verified.
(v) Answer :
Let p(x) = x2–15
To find zeroes, we can equate p(x) to 0.
So, p(x) = 0
By factoring, we get
(x+√15)(x-√15) = 0
Verifying whether they are zeroes :
x+√15 = 0 x = -√15 p(-√15) = (-√15+√15) (-√15-√15) p(-√15) = 0 |
x-√15 = 0 x = √15 p(√15) = (√15+√15) (√15-√15) p(√15) = 0 |
So, -√15 and √15 are zeroes of the given quadratic polynomial.
So, α = -√15 and β = √15
By comparing the quadratic equation with
ax2+bx+c = 0
a = 1, b = 0 and c = -15
Verifying the relationship :
The sum of the zeroes (α+β) :
-√15 + √15 = 0/1
0 = 0
The product of the zeroes αβ :
(-√15)(√15) = -15
-15 = -15
Thus, the basic relationship verified.
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