Question

Find the zeros of the polynomial 6y2- 7y+2 and verify the relationship between the zeros and coefficients of the polynomial .​

Answer 1

★ To find :

Roots of 6y² - 7y + 2

And to verify the relationship between zeros and coefficients of polynomial..

★ Solution :

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 ,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)

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

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

→ Sum of roots :

In our case :

b (coefficient of y) = -7

a (leading coefficient/coefficient of y²) = 6

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

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

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

L.H.S = R.H.S

★ Hence , verified .

→ Product of roots :

c (constant term) = 2

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

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

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

→ L.H.S = R.H.S

★ Hence, verified..

Answer 2

Answer:

Let the roots of the given equation be 'a' &'b' .

a+b=-(-7)/6=7/6. .....(1)

ab=2/6=1/3. .....(2)

Now,the required equation has roots: 1/a & 1/b.

sum of roots,S=(1/a)+(1/b)

=(a+b)/ab

=(7/6)÷(1/3). {from 1 & 2}

S =7/2. ...(3)

product of roots,P=(1/a)×(1/b)

=1/(ab)

=1÷(1/3)

P =3. ...(4)

The required equation will be of the form:

y^2 -Sy+P = y^2-(7/2)y+3=O [from 3 & 4]

=2y^2-7y+6=0

Hence,the required equation is:

2y^2 - 7y + 6 = 0

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Comments Report

Mathexpert

Mathexpert Expert

6y²-7y+2

6y²-4y-3y+2

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

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

Zeroes = 2/3 or 1/2

If alpha = 2/3 and beta = 1/2

then 1/alpha = 3/2 and 1/beta = 2

Now the quadratic polynomial is

x²-(sum of roots)x + (product of the roots)

Sum of the roots = 3/2 + 2 = 3.5

product of the roots = 3/2 * 2 = 3

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