Question

How can I find the zero of the quadratic polynomial and relation between the zeros and coefficient of 3x2-x-4

Answer 1

Answer:

Zeroes = 4/3 and (-1)

Step-by-step explanation:

Given quadratic polynomial :

• 3x² - x - 4

We can find the zeroes of the quadratic polynomial by splitting its middle term ;

$\sf \implies 3x^2 - x - 4 = 0 \\\\\implies \sf 3x^2 - 4x + 3x - 4 = 0 \\\\\implies \sf 3x^2 + 3x - 4x - 4 = 0\\\\\implies \sf 3x(x+1)-4(x+1) = 0 \\\\\implies \sf (3x-4)(x+1) = 0 \\\\\implies \sf x = \frac{4}{3} \: \: or \: \: x = - 1$

Thus, the zeroes of the quadratic polynomial is 4/3 and - 1.

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Now, to verify the relationship between the zeroes and the coefficient ;

We know that,

Sum of zeroes = $\sf \dfrac{-(coefficient \: of \: x)}{coefficient \: of \: x^2}$

$\sf \implies \frac{4}{3} - 1 = \frac{-(-1)}{3} \\\\\implies \sf \frac{4-3}{3} = \frac{1}{3} \\\\\implies \sf \frac{1}{3} = \frac{1}{3} \: \: \: \bf{[\therefore LHS = RHS]}$

Now,

Product of zeroes = $\sf \dfrac{Constant \: term}{coefficient \: of \: x^2}$

$\sf \implies \frac{4}{3} \times (- 1) = \frac{-4}{3}\\\\\implies \sf \frac{-4}{3} = \frac{-4}{3} \: \: \: \bf{[\therefore LHS = RHS]}$

Hence, verified!.

Answer 2

GIVEN:

Quadratic polynomial = 3x² - x - 4

TO FIND:

Zeroes of polynomial and verify relation between the zeros and coefficients.

SOLUTION:

To find the zeroes of polynomial we can use splitting the middle term method.

==> 3x² - x - 4 = 0

Split the middle term of the polynomial

==> 3x² + 3x - 4x - 4

==> 3x(x + 1) - 4(x + 1)

==> x + 1 = 0 ; 3x - 4 = 0

==> x = -1 ; x = 4/3

Now, verify the relation between the zeros and coefficient.

VERIFICATION:

Using zeroes:

Sum of zeores = -1 + 4/3

= 1/3

Product of zeroes = -1 × 4/3

= -4/3

Using Coefficients:

Sum of zeroes = -b/a

= -(-1)/3 = 1/3

Product of zeroes = c/a

= -4/3

Hence, verified!