Question

find the zeroes of the quadratic polynomial and verify their relation ships with the zeros of the polynomial 2x^2 + 5x - 12

Answer 1

Given :-

• Quadratic Equation is 2x² + 5x - 12 = 0

To Find :-

• Zeros of The polynomial and verify their relation ships with the zeros.. ?

Solution :-

→ 2x² + 5x - 12 = 0

Splitting The Middle Term we get,

→ 2x² + 8x - 3x - 12 = 0

→ 2x(x + 4) - 3(x + 4) = 0

→ (2x -3)(x + 4) = 0

Putting Both Equal to Zero now, we get,

→ 2x - 3 = 0

→ 2x = 3

→ x = (3/2)

Or,

→ x + 4 = 0

→ x = (-4).

Hence, Zeros of The polynomial are (-4) and (3/2).

____________________________

Now, First Relation is :-

→ Sum of Zeros = - (coefficient of x) /(coefficient of x²)

Putting both values ,

→ (3/2) + (-4) = -(5)/2

→ (3-8) /2= (-5)/2

→ (-5/2) = (-5/2) ✪✪ Hence Verified. ✪✪

Second Relation :-

→ Product Of Zeros = Constant Term / (coefficient of x²)

Putting both Values ,

→ (3/2) * (-4) = (-12) / (2)

→ (-6) = (-6) ✪✪ Hence Verified. ✪✪

___________________________

Answer 2

Answer:

⋆ Given Polynomial : 2x² + 5x – 12

Here : a = 2,⠀b = 5,⠀c = – 12

$:\implies\tt f(x) = 0\\\\\\:\implies\tt 2x^2 +5x-12 = 0\\\\\\:\implies\tt 2x^2+8x-3x-12 = 0\\\\\\:\implies\tt 2x(x+4)-3(x+4) = 0\\\\\\:\implies\tt (2x-3)(x+4)=0\\\\\\:\implies\underline{\boxed{\tt x =\dfrac{3}{2}\quad or \quad - \:4}}$

$\rule{160}{1}$

$\underline{\bigstar\:\textsf{Relation b/w zeroes and coefficient :}}$

$\qquad\underline{\bf{\dag}\:\:\textsf{Sum of Zeroes :}}\\\dashrightarrow\tt\:\: \alpha+\beta = \dfrac{- \:b}{a}\\\\\\\dashrightarrow\tt\:\: \dfrac{3}{2} +( - \:4) = \frac{ - \:5}{2}\\\\\\\dashrightarrow\tt\:\: \dfrac{(3 -8)}{2} = \dfrac{ - \:5}{2}\\\\\\\dashrightarrow\:\:\underline{\boxed{\red{\tt \dfrac{-\:5}{2} =\dfrac{-\:5}{2}}}} \\\\\\\qquad\underline{\bf{\dag}\:\:\textsf{Product of Zeroes :}}\\\\\dashrightarrow\tt\:\: \alpha \times \beta = \dfrac{c}{a}\\\\\\\dashrightarrow\tt\:\: \dfrac{3}{2} \times( - \:4) = \dfrac{ - \:12}{2}\\\\\\\dashrightarrow\tt\:\:3 \times( - \:2) = - \:6\\\\\\\dashrightarrow\:\:\underline{\boxed{ \red{\tt - \:6 = - \:6}}}$