Question

find the zeroes of the following quadratic polynomial and varify the relentonship between the zeroes and the corfficent : f(x)=6x^-3-7x
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Answer 1

AnswEr:-

Zeros of polynomial = -1/3 & 3/2

$\rule{200}{1}$

Here, given polynomial is :

☛ f(x) = 6x² - 3 - 7x = 0

We can find zeros of polynomial by factorization method:-

⇒ 6x² - 3 - 7x = 0

⇒ 6x² - 7x - 3 = 0

⇒ 6x² - 9x + 2x - 3 = 0

⇒ 3x(2x - 3) + 1(2x - 3) = 0

⇒ (3x + 1)(2x - 3) = 0

⇒ 3x = -1 or 2x = 3

⇒ x = -1/3 or x = 3/2

Therefore,

$\therefore\underline{\textsf{Zeros of polynomial = {\textbf{-1/3 \& 3/2 }}}}$

$\rule{200}{1}$

Verification:-

Here,

• α = -1/3
• β = 3/2

And,

• Coefficient of x²(a) = 6
• Coefficient of x(b) = -7
• Constant term(c) = -3

Relationship 1:-

☛ Sum of zeros = -b/a

⇒ -1/3 + 3/2 = -(-7)/6

⇒ (-2 + 9)/6 = 7/6

⇒ 7/6 = 7/6 [Hence verified!]

Relationship 2:-

☛ Product of zeros = c/a

⇒ -1/3 × 3/2 = -3/6

⇒ -1/2 = -1/2 [Hence verified!]

Therefore,

Relation between the zeros and coefficients are verified!

Answer 2

Given polynomial:-

$\bold{3x^2-7x-3}$

To find:

The zeros of the given polynomial and verify the relationship between the zeroes and the coefficient.

Solution:-

Zeros of the given polynomial:

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

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

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

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

In the polynomial, a = 6, b = (-7) & c = (-3).

Now, let 3/2 be α and -1/3 be β.

• α + β = -b/a

• α + β = -(-7)/6

• α + β = 7/6

• 3/2 + (-1/3) = (9-2)/6

• 3/2 + (-1/3) = 7/6

So, here the relation is verified.(sum of zeroes)

• αβ = c/a

• αβ = -3/6

• αβ = -1/2

• (3/2)*(-1/3) = [3*(-1)]/(2*3)

• (3/2)*(-1/3) = -3/6

• (3/2)*(-1/3) = -1/2

Here, again the relation is verified.(product of zeroes)

∴So, we can conclude that the relationship between the zeroes of the polynomial and the coefficients is verified.