Question

Find the zeros of the given quadratic polynomial and verify the relationship between the zeros

and the coefficients. 6x

2 + x – 12.​

Answer 1

EXPLANATION.

Quadratic equation = 6x² + x - 12

Sum of zeroes of quadratic equation

→ a + b = -b/a

→ a + b = -1/6.

Products of zeroes of quadratic equation

→ ab = c/a

→ ab = -12/6 = -2.

equation → 6x² + x - 12.

Factories into middle term split.

→ 6x² + 9x - 8x - 12 = 0.

→ 3x ( 2x + 3 ) - 4 ( 2x + 3 ) = 0.

→ ( 3x - 4 ) ( 2x + 3 ) = 0.

→ x = 4/3. and x = -3/2.

→ products = 4/3 X -3/2 = -2.

→ Sum = 4/3 + -3/2 = 8 - 9 / 6 = -1/6.

HENCE PROVED.

Answer 2

Given :-

• Quadratic Polynomial
• 6x² + x – 12.

To find :-

• Relationship between
• Sum of Zeroes of Polynomial
• Product of Zeroes of Polynomial

• Zeroes of the Polynomial

Concept :-

● Polynomial is always in the form of ax² + bx + c.

● Comparing the Polynomial 6x² + x – 12 with ax² + bx + c we get :-

• a = 6
• b = 1
• c = -12

● Sum of the Zeroes is represented by

${\underline{\boxed{\sf{Sum \: of \: the \: Zeroes = \alpha + \beta = - \frac{b}{a} = \frac{Coefficient \: of \: x}{Coefficient \: of \: {x}^{2} } }}}}$

● Product of Zeroes is represented by

${\underline{\boxed{\sf{Product \: of \: the \: Zeroes = \alpha \beta = \frac{c}{a} = \frac{Constant \: term}{Coefficient \: of \: {x}^{2} } }}}}$

Solution :-

■ Sum of the Zeroes is

${\sf{Sum \: of \: the \: Zeroes=-\cfrac{b}{a}}}$

${\sf{Sum \: of \: the \: Zeroes = - \cfrac{(1)}{(6)} }}$

${ \underline{ \overline{ \boxed{ \red{\bf{Sum \: of \: the \: Zeroes = - \cfrac{1}{6} }}}}}}$

■ Product of the Zeroes is

${\sf{Product \: of \: the \: Zeroes = \cfrac{ c}{a} }}$

${\sf{Product \: of \: the \: Zeroes = \cfrac{ - 12}{6} }}$

${ \underline{ \overline{ \boxed{ \red{\bf{Product \: of \: the \: Zeroes = -2 }}}}}}$

■ Zeroes or Roots of the Polynomial are :-

• For finding Zeroes we need to make Polynomial in equation form that will become 6x² + x – 12 = 0.

Factiorazing the 6x² + x - 12 by middle term splitting method :-

${\bf{\implies{6 {x}^{2} + x - 12 = 0}}}$

${\bf{\implies{6 {x}^{2} + 9x - 8x - 12 = 0}}}$

${\bf{\implies{3x(2x + 3) - 4(2x + 3) = 0}}}$

${\bf{\implies{(3x- 4)(2x + 3) = 0}}}$

■ So, Zeroes of the polynomial are :-

1st root

$\: \: \: \: \: \: \: \: \: \: {\bf{\implies{3x -4 = 0}}} \\ \: {\bf{\implies{3x = 4}}} \\ { \underline{ \overline{ \boxed{ \red{\bf{\implies{x = \frac{4}{3} }}}}}}}$

2nd root

$\: \: \: \: \: \:{\bf{\implies{2x + 3 = 0}}} \\ \: {\bf{\implies{2x = - 3}}} \\ { \underline{ \overline{ \boxed{ \red{\bf{\implies{x = \frac{ - 3}{2} }}}}}}}$