Question

write the sum of zeroes

3x2 -x-4​

Answer 1

GIVEN :-

• A quadratic polynomial 3x² - x - 4 = 0.

TO FIND :-

• The sum of zeroes [ ɑ + β ].

SOLUTION :-

Let us assume that,

⇒ ɑ = [ -b + √( b² - 4ac ) ] / 2a.

• a = 3.
• b = -1.
• c = -4.

⇒ ɑ = [-(-1) + √{(-1)² - 4 × 3 × (-4)}]/2 × 3

⇒ ɑ = [1 + √{1 - 12 × (-4)}]/6

⇒ ɑ = [1 + √{1 - (-48)}]/6

⇒ ɑ = [1 + √(1 + 48)]/6

⇒ ɑ = [1 + √49]/6

⇒ ɑ = (1 + 7)/6

⇒ ɑ = 8/6

⇒ ɑ = 4/3.

Now Similarly we will find the value of β,

⇒ β = [ -b - √( b² - 4ac ) ] / 2a.

• a = 3.
• b = -1.
• c = -4.

⇒ β = [-(-1) - √{(-1)² - 4 × 3 × (-4)}]/2 × 3

⇒ β = [1 - √{1 - 12 × (-4)}]/6

⇒ β = [1 - √{1 - (-48)}]/6

⇒ β = [1 - √(1 + 48)]/6

⇒ β = [1 - √49]/6

⇒ β = (1 - 7)/6

⇒ β = -6/6

⇒ β = -1.

Now according to question we have to find the value of ɑ + β , So,

⇒ ɑ + β = 4/3 + (-1)

⇒ ɑ + β = 4/3 - 1

⇒ ɑ + β = (4 - 3)/3

⇒ ɑ + β = 1/3

Hence the sum of the zeroes of the given quadratic polynomial is 1/3.

Alternative method :-

⇒ ɑ + β = (-coefficient of x)/(cofficient of x²)

⇒ ɑ + β = -b/a

⇒ ɑ + β = -(-1)/3

⇒ ɑ + β = 1/3

Hence the sum of the zeroes of the given quadratic polynomial is 1/3.

Answer 2

Given:-

• Quadratic eq. 3x² - x - 4 = 0

Find:-

• Sum of its zeroes.

Solution:-

Compare 3x² - x - 4 = 0 with ax² + bx + c = 0

So,

a = 3

b = -1

c = -4

Now, we know that

$\underline{\boxed{\sf Sum \: of \: zeroes = \dfrac{ - coefficient \: of \:x }{coefficient \: of \: {x}^{2} }}}$

$\implies\sf Sum \: of \: zeroes = \dfrac{ - b }{a}$

$\sf where \small{\begin{cases} \sf b = - 1 \\ \sf a = 3 \end{cases}}$

Substituting these values:-

$\dashrightarrow\sf Sum \: of \: zeroes = \dfrac{ - b }{a}$

$\dashrightarrow\sf Sum \: of \: zeroes = \dfrac{ - ( - 1)}{3}$

$\dashrightarrow\sf Sum \: of \: zeroes = \dfrac{ \not{-} ( \not{-} 1)}{3}$

$\dashrightarrow\sf Sum \: of \: zeroes = \dfrac{ 1}{3}$

$\small{\underline{\boxed{\sf\therefore Sum\:of\:zeroes\:is\:\dfrac{1}{3}}}}$