Question

find the zeros of the quadratic polynomial 6 x ^2 - 13 x + 6 and verify the relation between zeros and its coefficient. ​

Answer 1

Given

Quadratic polynomial 6x² - 13x + 6

Finding zeroes

6x² - 13x + 6

Splitting the middle term

= 6x² - 9x - 4x + 6

= 3x( 2x - 3 ) - 2( 2x - 3 )

= ( 3x - 2 )( 2x - 3 )

Zeroes

⇒ ( 3x - 2 )( 2x - 3 ) = 0

⇒ 3x - 2 = 0 or 2x - 3 = 0

⇒ x = 2 / 3 or x = 3 / 2

Therefore the zeroes of the polynomial are 2 / 3 and 3 / 2

Verification:

Comparing 6x² - 13x + 6 with ax² + bx + c we get,

• a = 6

• b = - 13

• c = 6

Let α, β be the zeroes of the polynomial

Sum of zeroes = α + β = - b / a

⇒ 2 / 3 + 3 / 2 = - ( - 13 ) / 6

⇒ ( 4 + 9 ) / 6 = 13 / 6

⇒ 13 / 6 = 13 / 6

Product of zeroes = αβ = c / a

⇒ ( 2 / 3 ) × ( 3 / 2 ) = 6 / 6

⇒ 1 = 1

LHS = RHS

Hence the relation is verified.

Answer 2

$\huge{ \underline{ \underline{ \boxed{ \sf{AnsweR :}}}}}$

• Given polynomial : 6x² - 13x + 6.

Here,

• a = 6
• b = -13
• c = 6

We know that the sum of zeroes = $\frac{ - b}{a}$

Which is $\frac{ 13}{6}$

And the product of zeroes = $\frac{c}{a}$

Which is $\frac{6}{6} = 1$

Now let's split the middle term of the given quadratic polynomial to get the value of x.

⟹ 6x² - 13x + 6

⟹ 6x² - 9x - 4x + 6

⟹ 3x ( 2x - 3 ) - 2 ( 2x - 3 )

⟹ ( 2x - 3 ) ( 3x - 2 )

• x = $\frac{3}{2}$ ... Let's take this value as $\alpha$

• x = $\frac{2}{3}$... And let us take this value as $\beta$

____________________________

Let us verify the relation between zeroes and coefficients.

The sum of zeroes =

$\alpha + \beta \: = \frac{3}{2} + \frac{2}{3}$

$\alpha + \beta = \frac{13}{6}$

And the product of zeroes =

$\alpha \times \beta = \frac{3}{2} \times \frac{2}{3}$

$\alpha \times \beta = 1$

$\sf{Hence \: Relation \: Verified!}$

Thank You! :)