Math, asked by hitvanshu, 11 months ago

find the zeroes of quadratic polynomial 7x^2-11/3x -2/3​

Answers

Answered by anshikakesari
18

7x^2-11/3x-2/3

dividing by 3 in equation

21x^2-11x-2

21x^2 -14x +3x -2

7x(3x-2) +1(3x-2)

(7x+1) (3x-2)

7x+1=0. ,3x-2=0

x =-1/7

x=2/3


anshikakesari: from my opinion it is not.. if any so I don't know
hitvanshu: ok
hitvanshu: thanks for help
anshikakesari: ys we can solve by not dividing it by 3... but it came some complicated
anshikakesari: your welcome
anshikakesari: if u want I can also solve without dividing it by 3
hitvanshu: how
anshikakesari: first multiple 7 and 2/3 it become 14/3 ... then add or subtract it and make... I will subtract and make 14/3 and -3/3
anshikakesari: and then take common
anshikakesari: and I think u will solve know
Answered by arshikhan8123
0

Concept:

The polynomial equations of degree two in one variable of type f(x) = ax2 + bx + c = 0 and with a, b, c, and R R and a 0 are known as quadratic equations. It is a quadratic equation in its general form, where "a" stands for the leading coefficient and "c" for the absolute term of f. (x). The roots of the quadratic equation are the values of x that fulfil the equation (α,β ).

It is a given that the quadratic equation has two roots. Roots might have either a true or imaginary nature.

Given:

7x²-11/3x-2/3=0

Find:

Find the zeroes of quadratic polynomial 7x^2-11/3x -2/3​

Solution:

7x²-11/3x-2/3=0

dividing by 3 in equation

21x²-11x-2=0

21x² -14x +3x -2=0

7x(3x-2) +1(3x-2)=0

(7x+1) (3x-2)=0

7x+1=0. ,3x-2=0

x =-1/7 x=2/3

Therefore, the roots of the equation are x =-1/7 x=2/3

#SPJ3

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