Question

Find the zeroes of the following quadratic polynomials and verify the relationship between
the zeroes and the coefficients.

(ii) 4s2 – 4s +1
(iii) 6x2 – 3-7x
(iv) 4u2 + 8u
(v) - 15
(vi) 3x2 - X-4​

Answer 1

Answer:

ii) sum of the zeroes=1, product of the zeroes=1/4,iii)sum of the zeroes=7/6, product of the zeroes=-3/6,iv)sum of the zeroes=

Step-by-step explanation:

ii)=4s^2-4s+1

=4s^2-2s-2s+1

=2s(2s-1)-1(2s-1)

=(2s-1)(2s-1)

Zeroes are 1/2 and 1/2

Let alpha=1/2,beta=1/2

Sum of zeroes=alpha+beta

=1/2+1/2

=1+1/2

=2/2=1

-b/a=-(-4)/4=1

-b/a= -( coefficient of x)/(coefficient of x^2)

Product of the zeroes= alpha× beta

=1/2×1/2

=1/4

C/a= 1/4

C/a= constant/(coefficient of x^2)

___________________________

Answer 2

$\huge\fbox \red{Solution:}$

(ii) g(s) = 4s2 – 4s + 1

Solution:

Given,

g(s) = 4s2 – 4s + 1

To find the zeros, we put g(s) = 0

⇒ 4s

2 – 4s + 1 = 0

⇒ 4s2

- 2s - 2s + 1= 0

⇒ 2s(2s - 1) - (2s - 1) = 0

⇒ (2s - 1)(2s – 1) = 0

This gives us 2 zeros, for

s = 1/2 and s = 1/2

Hence, the zeros of the quadratic equation are 1/2 and 1/2.

Now, for verification

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

Therefore, the relationship between zeros and their coefficients is verified.

(iii) h(t)=t2 – 15

Solution:

Given,

h(t) = t

2 – 15 = t2 +(0)t – 15

To find the zeros, we put h(t) = 0

⇒ t

2 – 15 = 0

⇒ (t + √15)(t - √15)= 0

This gives us 2 zeros, for

t = √15 and t = -√15

Hence, the zeros of the quadratic equation are √15 and -√15.

Now, for verification

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

Therefore, the relationship between zeros and their coefficients is verified.

(iv) f(x) = 6x2 – 3 – 7x

Solution:

Given,

f(x) = 6x

2 – 3 – 7x

To find the zeros, we put f(x) = 0

⇒ 6x

2 – 3 – 7x = 0

⇒ 6x

2

- 9x + 2x - 3 = 0

⇒ 3x(2x - 3) + 1(2x - 3) = 0

⇒ (2x - 3)(3x + 1) = 0

This gives us 2 zeros, for

x = 3/2 and x = -1/3

Hence, the zeros of the quadratic equation are 3/2 and -1/3.

Now, for verification

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

Therefore, the relationship between zeros and their coefficients is verified.

(v) p(x) = x

2 + 2√2x – 6

Solution:

Given,

p(x) = x2 + 2√2x – 6

To find the zeros, we put p(x) = 0

⇒ x

2 + 2√2x – 6 = 0

⇒ x

2 + 3√2x - √2x - 6 = 0

⇒ x(x + 3√2) - √2 (x + 3√2) = 0

⇒ (x - √2)(x + 3√2) = 0

This gives us 2 zeros, for

x = √2 and x = -3√2

Hence, the zeros of the quadratic equation are √2 and -3√2.

Now, for verification

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

Therefore, the relationship between zeros and their coefficients is verified.

(vi) q(x)=√3x2 + 10x + 7√3

Solution:

Given,

q(x) = √3x

2 + 10x + 7√3

To find the zeros, we put q(x) = 0

⇒ √3x

2 + 10x + 7√3 = 0

⇒ √3x

2 + 3x +7x + 7√3x = 0

⇒ √3x(x + √3) + 7 (x + √3) = 0

⇒ (x + √3)(√3x + 7) = 0

This gives us 2 zeros, for

x = -√3 and x = -7/√3

Hence, the zeros of the quadratic equation are -√3 and -7/√3.

Now, for verification

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

Therefore, the relationship between zeros and their coefficients is verified.