FIND ZEROES OF THE FOLLOWING QUADRATIC POLYNOMIAL AND VERIFY

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

Detailed Answer Key

(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 :

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 :

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 :

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 :

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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