Question

6. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes
and the coefficients.
6x2 – 3 – 7x
​

Answer 1

Given =》

6x²-3-7x

To Find =》

zeros =?

Solution =》

Arrange the term into general form

6x²-7x-3

6x²-(9-2)x-3

6x²-9x+2x-3

3x(2x-3)+1(2x-3)

(2x-3)(3x+1)

2x-3=0, 3x+1=0

2x=3, 3x=-1

x=3/2, x=-1/3

Zeros are 3/2 and -1/3

Sum of zeros = -b/a

$\frac{3}{2} + ( \frac{ - 1}{2} ) = \frac{ - ( - 7)}{6} \\ = \frac{3}{2} - \frac{1}{3} = \frac{7}{6} \\ = \frac{9 - 2}{6} = \frac{7}{6} \\ = \frac{7}{6} = \frac{7}{6}$

Product of zeros = c/a

$\frac{3}{2} \times ( \frac{ - 1}{3} ) = \frac{ - 3}{6} \\ = \frac{ - 3}{6} = \frac{ - 3}{6}$

Hence verified!

Answer 2

Step-by-step explanation:

Given =》

6x²-3-7x

To Find =》

zeros =?

Solution =》

Arrange the term into general form

6x²-7x-3

6x²-(9-2)x-3

6x²-9x+2x-3

3x(2x-3)+1(2x-3)

(2x-3)(3x+1)

2x-3=0, 3x+1=0

2x=3, 3x=-1

x=3/2, x=-1/3

Zeros are 3/2 and -1/3

Sum of zeros = -b/a

\begin{gathered}\frac{3}{2} + ( \frac{ - 1}{2} ) = \frac{ - ( - 7)}{6} \\ = \frac{3}{2} - \frac{1}{3} = \frac{7}{6} \\ = \frac{9 - 2}{6} = \frac{7}{6} \\ = \frac{7}{6} = \frac{7}{6}\end{gathered}

2

3

+(

2

−1

)=

6

−(−7)

=

2

3

−

3

1

=

6

7

=

6

9−2

=

6

7

=

6

7

=

6

7

Product of zeros = c/a

\begin{gathered}\frac{3}{2} \times ( \frac{ - 1}{3} ) = \frac{ - 3}{6} \\ = \frac{ - 3}{6} = \frac{ - 3}{6}\end{gathered}

2

3

×(

3

−1

)=

6

−3

=

6

−3

=

6

−3