Question

Find the zereos of the following quadratic polynomials and verify the relationship between the zereos and the co-efficients
1) 2x² - 9 - 3x
2) 2x² - 3 + 5x
3) 4u² + 8u
4) 3x² - x - 4

Answer 1

See the attach file for ur ans

Answer 2

Hi there !

Standard form of a quadratic polynomial :-

ax² + bx + y 
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sum of zeros of a polynomial = -b/a
product of zeros of a quadratic  polynomial = c/a
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1) 2x² - 9 - 3x
First of all arrange this is in standard form [ ax² + bx + y ]
2x² - 3x - 9
'Now , we can factorize this by splitting the middle term, equate it to zero .

2x² - 6x + 3x -9 = 0
2x [ x - 3 ] + 3 [ x - 3 ] = 0
[ 2x + 3 ] [ x - 3 ] = 0

2x + 3 = 0
x = -3/2

x - 3 = 0
x = 3

The zeros are 3 and -3/2.

2x² - 3x - 9 = 0
a = 2 , b = -3 , c = -9
-b/a = 3/2  , c/a = -9/2

sum of zeros = 3 + -3/2 = 6 - 3/2 = 3/2 = -b/a
product of zeros = 3 × -3/2 = -9/2 = c/a

Verified !
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2) 2x² - 3 + 5x
Arrange in standard form ,
 2x² + 5x - 3 = 0

factorize , by splitting the middle term ,
2x²+ 6x - x - 3 = 0
2x [ x + 3 ] - 1 [x + 3 ] = 0
[ 2x - 1 ] [ x + 3 ] = 0

2x - 1 = 0
x = 1/2

x + 3 = 0

x = -3

Therefore , -3 and 1/2 are zeros of the polynomial.
2x² + 5x - 3 = 0
a = 2 , b = 5 , c = -3

-b/a = -5/2 , 
c/a = -3/2

sum of zeros = -3 + 1/2 = -6+1/2 = -5/2 = -b/a
product of zeros = -3 × 1/2 = -3/2 = c/a

Verified !
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3)4u² + 8u

We can take 4u , which is common in both outside ,

4u [ u + 2 ] = 0
[ 4u + 0 ] [ u + 2 ] = 0

4u = 0
u = 0

u + 2 = 0
u = -2

∴ 0 and -2 are zeros of the polynomial.

4u² + 8u = 0

a = 4 , b = 8 , c = 0

-b/a = -8/4 = -2
c/a = 0/4 = 0

sum of zeros = 0 + -2 = -2 = b/a
product of zeros = 0 × -2 = 0 = c/a

Verified !
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4) 3x² - x - 4 

3x² + 3x - 4x - 4 = 0

3x [ x + 1 ] - 4 [ x + 1 ] = 0

[ 3x - 4 ] [ x + 1 ] = 0

3x - 4 = 0
x = 4/3

x + 1 = 0
x = -1

∴ -1 and 4/3 are zeros of the polynomial

 3x² - x - 4  = 0
a = 3 , b = -1 , c = -4

-b/a = -(-1)/3 = 1/3
c/a = -4/3

sum of zeros = -1 + 4/3 = -3+4/3 = 1/3 = -b/a
product of zeros = -1 × 4/3 = -4/3 = c/a

Verified !