Find the zeros of each of the following quadratic polynomials and verify the relationship between the zeros and their coefficients:
(i) f(x) = x² - 2x -8
(ii) g(s) = 4s² - 4s + 1
(iii) h(t) = t² - 15
(iv) 6x² - 3 - 7x
(v) p(x) = x² + 2√(2x) - 6
(vi) q (x) = √3 x² + 10x + 7√3
(vii) f(x) = x² - (√3 + 1)x + √3
(viii) g(x)=a(x² + 1) - x(a² + 1)
Answers
Solved and verified
Step-by-step explanation:
(i) f(x) = x² - 2x -8
zeroes of the polynomial is the value of x where p(x) = 0
x² - 2x -8 = 0
x² - 4x + 2x -8 = 0
x (x-4) + 2 (x-4) = 0
(x-4) (x+2) = 0
So x = -2, 4
α and β are the zeroes of the polynomial such that α = -2 and β = 4
α + β = -b/a
-2 +4 = -(-2)/1
2 = 2. Verified.
αβ = c/a
-2(4) = -8/1
-8 = -8. Verified.
Relationship between the zeroes and co-efficient is verified.
(ii) g(s) = 4s² - 4s + 1
4s² - 4s + 1 = 0
4s² - 2s - 2s + 1 = 0
2s (2s - 1) -1 (2s - 1) = 0
(2s-1) (2s-1) = 0
So s = +1/2, +1/2
α = β = +1/2
α + β = -b/a
+1/2 +1/2 = +4 / 4
+1 = +1
αβ = c/a
1/2 * 1/2 = 1/4
1/4 = 1/4
Relationship between the zeroes and co-efficient is verified.
(iii) h(t) = t² - 15
t² - 15 = 0
t² = 15
t = +-√15
α and β = +√15 and - √15
α + β = -b/a
+√15 -√15 = 0/1
0= 0
αβ = c/a
+√15 * -√15 = -15/1
-15 = -15
Relationship between the zeroes and co-efficient is verified.
(iv) 6x² - 3 - 7x
6x² - 7x -3 = 0
6x² - 9x +2x - 3 = 0
3x (2x -3) +1 (2x-3) = 0
α = +3/2 ; β = +1/3
(v) p(x) = x² + 2√(2x) - 6
x² + 2√(2x) - 6 = 0
x² + 3√2x -√2x -6 = 0
x (x + 3√2) -√2(x + 3√2) = 0
α = -3√2 ; β = +√2
(vi) q (x) = √3 x² + 10x + 7√3
√3 x² + 10x + 7√3 = 0
√3 x² + 10x + 7√3 = 0
√3 x² + 3x + 7x + 7√3 = 0
√3x (x +√3) + 7(x +√3) = 0
α = -√3 ; β = -7/√3
(vii) f(x) = x² - (√3 + 1)x + √3
x² - (√3 + 1)x + √3 = 0
x² - x -√3x + √3 = 0
x (x-1) -√3 (x - 1) = 0
α = 1 ; β = √3
(viii) g(x)=a(x² + 1) - x(a² + 1)
ax² + a - a²x - x = 0
ax² - (a²-1)x + a = 0
ax² -a²x - x + a = 0
ax (x - a) - 1 (x -a) = 0
(ax -1) (x-a)
α = a ; β = 1/a
Answer:
Given expresion is f(x) = 6x^2 − 3 − 7x
=> f(x) = 6x²-7x-3
=> f(x) = 6x²+2x-9x-3
=> f(x) = 2x(3x+1)-3(3x+1)
=> f(x) = (3x+1)(2x-3)
Factorization of f(x) = (3x+1)(2x-3)
To get zeroes we write f(x) = 0
=> (3x+1)(2x-3)=0
=> 3x+1 = 0 or 2x-3 = 0
=> 3x = -1 or 2x = 3
=> x = -1/3 or x = 3/2
Zeroes are -1/3 and 3/2
e)
Given expresion is p(x) = x^2 + 2√2x − 6
=> p(x) = x² + 2√2x − 6
=> p(x) = x²+3√2x-√2x - 6
=> p(x) = x(x+3√2)-√2(x+3√2)
=> p(x) = (x+3√2)(x-√2)
Factorization of p(x) = (x+3√2)(x-√2)
To get zeroes we write p(x) = 0
=> (x+3√2)(x-√2)=0
=> x+3√2 = 0 or x-√2= 0
=> x = -3√2 or x = √2
Zeroes are -3√2 and √2
f)
Given expresion is q(x) = √3x^2 + 10x + 7√3
=> q(x) =√3x²+10x+7√3
=> q(x) =√3x²+3x+7x+7√3
=> q(x) =√3x(x+√3)+7(x+√3)
=> q(x) = (x+√3)(√3x+7)
Factorization of q(x) = (x+√3)(√3x+7)
To get zeroes we write q(x) = 0
=> (x+√3)(√3x+7) = 0
=> x+√3 = 0 or √3x+7 = 0
=> x = -√3 or √3x = -7
=> x = -√3 or x = -7/√3
Zeroes are -√3 and -7/√3
g)
Given expresion is f(x) = x^2 − (√3 + 1)x + √3
=> f(x) = x²-√3x-x+√3
=> f(x) = x²-x-√3x+√3
=> f(x) = x(x-1)-√3(x-1)
=> f(x) = (x-1)(x-√3)
Factorization of f(x) = (x-1)(x-√3)
To get zeroes we write f(x) = 0
=> (x-1)(x-√3) = 0
=> x-1 = 0 or x-√3= 0
=> x = 1 or x = √3
Zeroes are 1 and √3
h)
Given expresion is g(x) = a(x^2 + 1) − x(a^2 + 1
=> g(x) = a(x² + 1) − x(a² + 1)
=> g(x) = ax²+a-a²x-x
=> g(x) = ax²-a²x-x+a
=> g(x)= ax(x-a)-1(x-a)
=> g(x) = (x-a)(ax-1)
Factorization of g(x) = (x-a)(ax-1)
To get zeroes we write g(x) = 0
=> (x-a)(ax-1) = 0
=>x-a = 0 or ax-1 = 0
=> x = a or ax = 1
=> x = 1 or x = 1/a
Zeroes are 1 and 1/a
Step-by-step explanation: