Question

And bare the two zeros of the polynomial 6y2-7y+2, find a quadratic polynomial i whose zeros are pue

Answer 1

Answer:

buddy I think your question is incomplete.

Complete It first, only then we would be able to solve it.

#be brainly

Answer 2

Answer:

The roots of the polynomial are - 2/3 and 1/2.

Step-by-step-explanation:

By middle term splitting

6y² - 7y + 2

= 6y² - 3y - 4y + 2

= 3y(2y - 1) -2 (2y - 1)

= (3y - 2)(2y-1)

Roots:

→ 3y - 2 = 0

3y = 2

y = 2/3

→ 2y - 1 = 0

2y = 1

y = 1/2

Therefore,

The roots of the polynomial are - 2/3 and 1/2.

To verify the relationship between zeros and coefficients of the polynomial,

We know that, (in a quadratic polynomial)

$\sf\leadsto Sum\: of \: roots = \frac{-b}{a}$

$\sf\leadsto Product\: of \:roots = \frac{c}{a}$

→ Sum of roots :

In this case :

b (coefficient of y) = -7

a (leading coefficient/coefficient of y²) = 6

$\sf\frac{2}{3} + \frac{1}{2} = \frac{-(-7)}{6}$

$\sf\frac{4 + 3}{6} = \frac{7}{6}$

$\sf\frac{7}{6} = \frac{7}{6}$

L.H.S = R.H.S

Hence verified!

→ Product of roots :

c (constant term) = 2

$\sf\frac{2}{3} \times \frac{1}{2} = \frac{\cancel{2}}{\cancel{6}}$

$\sf\frac{\cancel{2} \times 1}{3 \times \cancel{2}} = \frac{1}{3}$

$\sf\frac{1}{3} = \frac{1}{3}$

→ L.H.S = R.H.S

Hence verified!