Question

6x²- 3 -7x find the zeros and verify relation

Answer 1

Answer:

answer

$x = \frac{ - 1}{2} \\ x = \frac{3}{2}$

Step-by-step explanation:

$6 {x}^{2} - 3 - 7x = 0 \\ 6 {x}^{2} - 7x - 3 = 0 \\ 6 {x}^{2} - 9x + 2x - 3 = 0 \\ 3x(2x - 3) + 1(2x - 3) = 0 \\ 3x +1 = 0 \\ x = \frac{ - 1}{3} \\ 2x - 3 = 0 \\ x = \frac{3}{2}$

Answer 2

QUESTION:

6x²- 3 -7x find the zeros and verify relation

TO FIND :

☆zeroes

☆verify the relation between zeroes.

ANSWER:

As we know that.;quadratic equation is in the form of

$\red {a {x}^{2} + bx + c = 0}$

where;

b = sum of zeroes

c = product of zeroes

☆ In order to find the zeroes we have to factorised it.

Splitting the middle term :

Split the middle term in such a way that it's sum equal to b i.e -7 and product equal to a ×c = -18.

$6 {x}^{2} - 9x + 2x - 3 = 0$

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

$\huge\orange {(3x - 1)(2x - 3) = 0}$

so,

$3x 1 1 = 0 \\ 3x = -1 \\ x = \frac{-1}{3}$

$\huge\blue {x = \frac{-1}{3}}$

or

$2x - 3 = 0 \\ 2x = 3$

$\huge\pink {x = \frac{3}{2}}$

Relation:

product of zeroes =

$\frac{ - 1}{3} \times \frac{3}{2} \\ \frac{ -1}{2}$

we know that;

$\purple {product \: of \: zeroes = \frac{coefficient \: of \: x}{coefficient \: of \: {x}^{2} } = \frac{c}{a}}$

$product \: of \: zeroes = \frac{ - 3}{6} \\ = \frac{ - 1}{2}$

it is verified.

now,

sum of zeroes =

$\frac{ - 1}{3} + \frac{3}{2} \\ \frac{ - 2 + 9}{3 \times 2} \\ \frac{ 7}{6} \\ \frac{1}{3}$

we know that;

$\red {sum \: of \: zeroes = \frac{ - coefficient \: of \: x}{coeffiecint \: of \: {x}^{2} } = \frac{ - b}{a}}$

so,

$sum \: of \: zeroes = \frac{ - ( - 7)}{6} = \frac{7}{6}$

it is verified .