Question

Find the quadratic polynomial 2x^2-3x-9 and verify the relationship between the zeros and the coefficient

Answer 1

Step-by-step explanation:

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Given :

• Given quadratic polynomial is 2x²-3x-9

To Find :

• The relationship between the zeros and the coefficient.

Solution:

Quadratic polynomial 2x²-3x-9

On comparing with general formula

• a = 2
• b = -3
• c = -9

Splitting middle term,

$\implies \bf \: {2x}^{2} - 3x - 9$

$\implies \bf \: {2x}^{2} - 6x + 3x - 9 = 0$

$\implies \bf \:2x(x - 3) + 3(x - 3) = 0$

$\implies \bf \:(2x + 3)(x - 3) = 0$

$\implies \bf \:x = - \frac{3}{2} \: \: and \: \: 3$

Zeroes of the equation is -3/2 and 3

Relationship between the zeros and the coefficient :

$\sf \: Sum \: \: of \: \: zeroes \: ( \alpha + \beta ) = \frac{ - b}{a}$

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

$\frac{ - 3 + 6}{2} = \frac{3}{2}$

$\frac{3}{2} = \frac{3}{2}$

$\sf \: Product \: \: of \: \: zeroes \: ( \alpha \beta ) = \frac{c}{a}$

$\frac{ - 3}{2} \times 3 = \frac{ -9 }{2}$

$\frac{ - 9}{2} = \frac{ - 9}{2}$

Hence Verified !!

Answer 2

Answer:

➙ Question :-

Find the quadratic polynomial 2x^2-3x-9 and verify the relationship between the zeros and the coefficient

➙ Required Answer :-

➙ Explanation :-

we have

$2x^2-3x- \: 9$

Standard form :-

ax² + bx + c

Here the value of

a = 2

b = -3

c = -9

➙ Finding zeroes :-

We will find the zeroes of the polynomial by splitting the middle term

$2x^2-3x- \: 9$

$= 2x^2 - 6x + 3x-9$

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

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

So the value of p(x) = 2x²- 3x -9 is zero when 2x+3=0 or x-3 = 0

so , the zeros of 2x²- 3x -9 are -3/2 and 3.

➙ Verifying the relationship between coefficient and zero :-

Now let one of the zero be alpha and other beta

➙ Sum of zeroes :-

$\alpha + \beta = \frac{ - ( - b)}{a}$

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

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

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

$\frac{ - 3}{2} + \frac{6}{2} = \frac{3}{2}$

$\frac{ - 3 + 6}{2} = \frac{3}{2}$

$\frac{3}{2} = \frac{3}{2}$

➙ Product of zeroes :-

$\alpha \beta = \frac{c}{a}$

$\frac{ - 3}{2} \times 3 = \frac{ - 9}{2}$

$\frac{ - 3}{2} \times \frac{3}{1} = \frac{ - 9}{2}$

$\frac{ - 9}{2} = \frac{ - 9}{2}$

Since the values are same and hence verified.