Question 1. 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
Solution:
Given that,
f(x) = x2 – 2x – 8
To find the zeros of the equation, put f(x) = 0
= x2 – 2x – 8 = 0
= x2 – 4x + 2x – 8 = 0
= x(x – 4) + 2(x – 4) = 0
= (x – 4)(x + 2) = 0
x = 4 and x = -2
Hence, the zeros of the quadratic equation are 4 and -2.
Now, Verification
As we know that,
Sum of zeros = – coefficient of x / coefficient of x^2
4 + (-2)= – (-2) / 1
2 = 2
Product of roots = constant / coefficient of x^2
4 x (-2) = (-8) / 1
-8 = -8
Hence the relationship between zeros and their coefficients are verified.
(ii) g(s) = 4s2 – 4s + 1
Solution:
Given that,
g(s) = 4s2 – 4s + 1
To find the zeros of the equation, put g(s) = 0
= 4s2 – 4s + 1 = 0
= 4s2 – 2s – 2s + 1= 0
= 2s(2s – 1) – (2s – 1) = 0
= (2s – 1)(2s – 1) = 0
s = 1/2 and s = 1/2
Hence, the zeros of the quadratic equation are 1/2 and 1/2.
Now, Verification
As we know that,
Sum of zeros = – coefficient of s / coefficient of s2
1/2 + 1/2 = – (-4) / 4
1 = 1
Product of roots = constant / coefficient of s2
1/2 x 1/2 = 1/4
1/4 = 1/4
Hence the relationship between zeros and their coefficients are verified.
(iii) h(t)=t2 – 15
Solution:
Given that,
h(t) = t2 – 15 = t2 +(0)t – 15
To find the zeros of the equation, put h(t) = 0
= t2 – 15 = 0
= (t + √15)(t – √15)= 0
t = √15 and t = -√15
Hence, the zeros of the quadratic equation are √15 and -√15.
Now, Verification
As we know that,
Sum of zeros = – coefficient of t / coefficient of t2
√15 + (-√15) = – (0) / 1
0 = 0
Product of roots = constant / coefficient of t2
√15 x (-√15) = -15/1
-15 = -15
Hence the relationship between zeros and their coefficients are verified.
(iv) f(x) = 6x2 – 3 – 7x
Solution:
Given that,
f(x) = 6x2 – 3 – 7x
To find the zeros of the equation, we put f(x) = 0
= 6x2 – 3 – 7x = 0
= 6x2 – 9x + 2x – 3 = 0
= 3x(2x – 3) + 1(2x – 3) = 0
= (2x – 3)(3x + 1) = 0
x = 3/2 and x = -1/3
Hence, the zeros of the quadratic equation are 3/2 and -1/3.
Now, Verification
As we know that,
Sum of zeros = – coefficient of x / coefficient of x2
3/2 + (-1/3) = – (-7) / 6
7/6 = 7/6
Product of roots = constant / coefficient of x2
3/2 x (-1/3) = (-3) / 6
-1/2 = -1/2
Hence the relationship between zeros and their coefficients are verified.
(v) p(x) = x2 + 2√2x – 6
Solution:
Given that,
p(x) = x2 + 2√2x – 6
To find the zeros of the equation, put p(x) = 0
= x2 + 2√2x – 6 = 0
= x2 + 3√2x – √2x – 6 = 0
= x(x + 3√2) – √2 (x + 3√2) = 0
= (x – √2)(x + 3√2) = 0
x = √2 and x = -3√2
Hence, the zeros of the quadratic equation are √2 and -3√2.
Now, Verification
As we know that,
Sum of zeros = – coefficient of x / coefficient of x2
√2 + (-3√2) = – (2√2) / 1
-2√2 = -2√2
Product of roots = constant / coefficient of x2
√2 x (-3√2) = (-6) / 2√2
-3 x 2 = -6/1
-6 = -6
Hence the relationship between zeros and their coefficients are verified.
(vi) q(x)=√3x2 + 10x + 7√3
Solution:
Given that,
q(x) = √3x2 + 10x + 7√3
To find the zeros of the equation, put q(x) = 0
= √3x2 + 10x + 7√3 = 0
= √3x2 + 3x +7x + 7√3x = 0
= √3x(x + √3) + 7 (x + √3) = 0
= (x + √3)(√3x + 7) = 0
x = -√3 and x = -7/√3
Hence, the zeros of the quadratic equation are -√3 and -7/√3.
Now, Verification
As we know that,
Sum of zeros = – coefficient of x / coefficient of x2
-√3 + (-7/√3) = – (10) /√3
(-3-7)/ √3 = -10/√3
-10/ √3 = -10/√3
Product of roots = constant / coefficient of x2
(-√3) x (-7/√3) = (7√3) / √3
7 = 7
Hence the relationship between zeros and their coefficients are verified.
(vii) f(x) = x2 – (√3 + 1)x + √3
Solution:
Given that,
f(x) = x2 – (√3 + 1)x + √3
To find the zeros of the equation, put f(x) = 0
= x2 – (√3 + 1)x + √3 = 0
= x2 – √3x – x + √3 = 0
= x(x – √3) – 1 (x – √3) = 0
= (x – √3)(x – 1) = 0
x = √3 and x = 1
Hence, the zeros of the quadratic equation are √3 and 1.
Now, Verification
Sum of zeros = – coefficient of x / coefficient of x2
√3 + 1 = – (-(√3 +1)) / 1
√3 + 1 = √3 +1
Product of roots = constant / coefficient of x2
1 x √3 = √3 / 1
√3 = √3
Hence the relationship between zeros and their coefficients are verified.
(viii) g(x) = a(x2+1)–x(a2+1)
Solution:
Given that,
g(x) = a(x2+1)–x(a2+1)
To find the zeros of the equation put g(x) = 0
= a(x2+1)–x(a2+1) = 0
= ax2 + a − a2x – x = 0
= ax2 − a2x – x + a = 0
= ax(x − a) − 1(x – a) = 0
= (x – a)(ax – 1) = 0
x = a and x = 1/a
Hence, the zeros of the quadratic equation are a and 1/a.
Now, Verification :
As we know that,
Sum of zeros = – coefficient of x / coefficient of x2
a + 1/a = – (-(a2 + 1)) / a
(a^2 + 1)/a = (a2 + 1)/a
Product of roots = constant / coefficient of x2
a x 1/a = a / a
1 = 1
Hence the relationship between zeros and their coefficients are verified.
(ix) h(s) = 2s2 – (1 + 2√2)s + √2
Solution:
Given that,
h(s) = 2s2 – (1 + 2√2)s + √2
To find the zeros of the equation put h(s) = 0
= 2s2 – (1 + 2√2)s + √2 = 0
= 2s2 – 2√2s – s + √2 = 0
= 2s(s – √2) -1(s – √2) = 0
= (2s – 1)(s – √2) = 0
x = √2 and x = 1/2
Hence, the zeros of the quadratic equation are √3 and 1.
Now, Verification
As we know that,
Sum of zeros = – coefficient of s / coefficient of s2
√2 + 1/2 = – (-(1 + 2√2)) / 2
(2√2 + 1)/2 = (2√2 +1)/2
Product of roots = constant / coefficient of s2
1/2 x √2 = √2 / 2
√2 / 2 = √2 / 2
Hence the relationship between zeros and their coefficients are verified.
(x) f(v) = v2 + 4√3v – 15
Solution:
Given that,
f(v) = v2 + 4√3v – 15
To find the zeros of the equation put f(v) = 0
= v2 + 4√3v – 15 = 0
= v2 + 5√3v – √3v – 15 = 0
= v(v + 5√3) – √3 (v + 5√3) = 0
= (v – √3)(v + 5√3) = 0
v = √3 and v = -5√3
Hence, the zeros of the quadratic equation are √3 and -5√3.
Now, for verification
Sum of zeros = – coefficient of v / coefficient of v2
√3 + (-5√3) = – (4√3) / 1
-4√3 = -4√3
Product of roots = constant / coefficient of v2
√3 x (-5√3) = (-15) / 1
-5 x 3 = -15
-15 = -15
Hence the relationship between zeros and their coefficients are verified.
(xi) p(y) = y2 + (3√5/2)y – 5
Solution:
Given that,
p(y) = y2 + (3√5/2)y – 5
To find the zeros of the equation put f(v) = 0
= y2 + (3√5/2)y – 5 = 0
= y2 – √5/2 y + 2√5y – 5 = 0
= y(y – √5/2) + 2√5 (y – √5/2) = 0
= (y + 2√5)(y – √5/2) = 0
This gives us 2 zeros,
y = √5/2 and y = -2√5
Hence, the zeros of the quadratic equation are √5/2 and -2√5.
Now, Verification
As we know that,
Sum of zeros = – coefficient of y / coefficient of y2
√5/2 + (-2√5) = – (3√5/2) / 1
-3√5/2 = -3√5/2
Product of roots = constant / coefficient of y2
√5/2 x (-2√5) = (-5) / 1
– (√5)2 = -5
-5 = -5
Hence the relationship between zeros and their coefficients are verified.
(xii) q(y) = 7y2 – (11/3)y – 2/3
Solution:
Given that,
q(y) = 7y2 – (11/3)y – 2/3
To find the zeros of the equation put q(y) = 0
= 7y2 – (11/3)y – 2/3 = 0
= (21y2 – 11y -2)/3 = 0
= 21y2 – 11y – 2 = 0
= 21y2 – 14y + 3y – 2 = 0
= 7y(3y – 2) – 1(3y + 2) = 0
= (3y – 2)(7y + 1) = 0
y = 2/3 and y = -1/7
Hence, the zeros of the quadratic equation are 2/3 and -1/7.
Now, Verification
As we know that,
Sum of zeros = – coefficient of y / coefficient of y2
2/3 + (-1/7) = – (-11/3) / 7
-11/21 = -11/21
Product of roots = constant / coefficient of y2
2/3 x (-1/7) = (-2/3) / 7
– 2/21 = -2/21
Hence the relationship between zeros and their coefficients are verified.
Question 2. For each of the following, find a quadratic polynomial whose sum and product respectively of the zeros are as given. Also, find the zeros of these polynomials by factorization.
(i) -8/3, 4/3
Solution:
As we know that the quadratic polynomial formed for the given sum and product of zeros is given by : f(x) = x2 + -(sum of zeros) x + (product of roots)