Find the zeros of the quadratic polynomials and verify the relationships between the zeros and the coefficients.

(i) x²-2x-8

Answer:

Let, P(x) = x²-2x-8

Comparing P(x) with a quadratic equation ax²+bx+c

we have b = -2 and c = -8.

P(x) = x²-2x-8

= x²-4x+2x-8

= x(x-4) + 2(x-4)

= (x-4) (x+2)

Therefore, the value of P(x) will be zero if x-4=0 or x+2=0 i.e., when x=4 or x=-2

Hence the zeros of P(x) are 4, -2

Sum of zeros = 4+(-2) = 2/1 = –-b/a = (-Coefficient of x)/(Coefficient of x2)

Product of the zeros = 4(-2) = -8 = -8/1 = c/a = Constant term / Coefficient of x2

(ii) 4s²-4s+1

Answer:

Let P(x) = 4s²-4s+1

To find the zeros of the quadratic polynomial we consider

4s² − 4s + 1 = 0

4s² − 2s – 2s + 1 = 0

2s(2s – 1) -1(2s – 1) = 0

(2s – 1)(2s – 1) = 0

2s – 1 = 0 and 2s – 1 = 0

2s = 1 and 2s = 1

s = 1/2 and s = 1/2

∴ The zeroes of the polynomial = 1/2 and 1/2

Sum of the zeroes = -(coefficient of s)/(coefficient of s²)

Sum of the zeroes = -(-4)/4 = 1

Let’s find the sum of the roots = 1/2 + 1/2 = 1

Product of the zeros = Constant term / Coefficient of s²

Product of the zeros =1 / 4

Let’s find the products of the roots = 1/2 × 1/2 = 1/4

(iii) 6x²-3-7x

Answer:

Let, P(x) = 6x²-3-7x

To find the zeros

Let us put f(x) = 0

⇒ 6x² – 7x – 3 = 0

⇒ 6x² – 9x + 2x – 3 = 0

⇒ 3x(2x – 3) + 1(2x – 3) = 0

⇒ (2x – 3)(3x + 1) = 0

⇒ 2x – 3 = 0

x = 3/2

⇒ 3x + 1 = 0

⇒ x = -1/3

It 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.

(iv) 4u²+8u

Answer:

Let, P(u) = 4u²+8u

Comparing P(u) with a quadratic equation au²+bu+c we have a = 4 ,b= 8  and c = 0

P(u) = 4u²+8u

   = u(4u+8)

Therefore, the value of P(u) will be zero

Then, u = 0

Or, 4u+8 = 0

4u = -8

u = -2

Hence the zeros of P(s) are 0, -2

Now sum of zeros = 0+(-2) =-2 = -2

And product of the zeros = 0 X 2= 0

But, the Sum of the zeroes in any quadratic polynomial equation is given by = −coeff.of u / coeff.of u² = −8/4 = −2

and, the product of zeroes in any quadratic polynomial equation is given by = constant term / coeff.of u² = −0/4 = 0

(v) t²-15

Answer:

Let P(t) = t²-15

Comparing P(t) with a quadratic equation at²+bt+c we have a = 1,b= 0 and c = -15

Therefore, the value of P(t) will be zero if t²-15 = 0 i.e., t²= 15 so t = ±√15

Hence the zeros of P(s) are √15, – √15

Sum of zeros = √15 – √15 = 0 = (−coeff.of t / coeff.of t²)

Product of the zeros = √15(-√15) = -15 =(constant term / coeff.of t²)

(vi) 3x²– x-4

Answer:

Let P(x) = 3x²-x-4

Comparing P(x) with a quadratic equation ax²+bx+c we have a = 3, b= -1 and c = –4

P(x) = 3x²-x-4

= 3x²-4x+3x-4

= x(3x-4)+1(3x-4)

= (3x-4)(x+1)

Therefore the value of P(x) will be zero if 3x-4=0 or x+1=0

Then, 3x = 4  

x = 4/3

Or, x = -1

Hence the zeros of P(x) are 4/3, – 1

Sum of zeros = 4/3 – 1= 1/3  = -b/a = (-Coefficient of x)/(Cofficient of x²)

Product of the zeros = (4/3) (-1) = –4/3  = = (constant term)/(Cofficient of x²)