Question

the zeros of the polynomial X^2 + 1/6 X - 2 are ------»
___________________
solve it with full solution please......
__________________
❎❎NO SPAMMING ❎ ❎
___________________​

Answer 1

Given :-

A polynomial $\sf x^2+\dfrac{1}{6}x-2=0$

To find :-

The zeroes of polynomial

• As we know that !

• By middle split term we can solve this

$\dashrightarrow\sf x^2+\dfrac{1}{6}x-2=0\\ \\ \sf \ \ \ By \ taking \ L.C.M\\ \\ \dashrightarrow\sf \dfrac{(6x^2+x-12)}{6}=0\\ \\ \dashrightarrow\sf 6x^2+x-12=6\times 0\\ \\ \dashrightarrow\sf 6x^2+(9-8)x-12=0\\ \\ \dashrightarrow\sf 6x^2+9x-8x-12=0\\ \\ \dashrightarrow\sf 3x(2x+3)-4(2x+3)=0\\ \\ \dashrightarrow\sf (3x-4)(2x+3)=0$

• Now the zeroes of polynomial !

● The zeroes of polynomial is point where the polynomial becomes zero !

• Now we have

$\dashrightarrow\sf (3x-4)(2x+3)=0\\ \\ \sf (i) 3x-4=0 \\ \\ \dashrightarrow\sf 3x=4\\ \\ \dashrightarrow\sf x=\dfrac{4}{3}\\ \\ \sf (ii) 2x+3=0\\ \\ \dashrightarrow\sf 2x=-3 \\ \\ \dashrightarrow\sf x=\dfrac{-3}{2}$

$\bigstar{\boxed{\sf{zeroes\ of \ polynomial = \dfrac{4}{3}\ \ \ and \ \dfrac{-3}{2}}}}$

Answer 2

Given

Quadratic equation : x² + (1/6)x - 2

To find

Zeros of polynomial

Solution

Let's start with the given polynomial :

➻ x² + (1/6)x - 2 = 0

➻ x² + (x/6) - 2 = 0

➻ (6x² + x - 12)/6 = 0

➻ 6x² + x - 12 = 0

➻ 6x² + 9x - 8x - 12 = 0

➻ 3x(2x + 3) - 4(2x + 3) = 0

➻ (3x - 4) (2x + 3) = 0

Now putting both equal to zero we get :

 ➊ 3x - 4 = 0

➪ 3x = 4

➪ x = 4/3

 ➋ 2x + 3 = 0

➯ 2x = -3

➯ x = -3/2

Tʜᴇʀᴇꜰᴏʀᴇ,

Zeros of polynomial are: 4/3 & (-3/2)