Question

find the zeros of the quadratic polynomial 2x square -9 - 3x and verify the relationship between the zeros and the coefficient

Answer 1

Answer:

The zeros of the equation are $x=-\frac{3}{2},3$

Step-by-step explanation:

Given : Equation $2x^2-9-3x$

To find : The zeros of the quadratic polynomial and verify the relationship between the zeros and the coefficient ?

Solution :

Quadratic equation $2x^2-3x-9=0$

On comparing with general equation, a=2,b=-3,c=-9

Applying middle term split,

$2x^2-6x+3x-9=0$

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

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

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

$x=-\frac{3}{2},x=3$

The zeros of the equation are $x=-\frac{3}{2},3$

The relationship between the zeros and the coefficient,

Sum of the zeros, $\alpha+\beta=-\frac{b}{a}$

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

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

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

Verified.

Product of the zeros, $\alpha\beta=\frac{c}{a}$

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

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

Verified.

Answer 2

Answer:

$-3 \:and \:\frac{3}{2}\:are \\ zeroes \:of \:given \: polynomial.$

Step-by-step explanation:

We have 2x²-3x-9

= 2x²+6x-3x-9

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

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

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

i.e ., when x = -3 Or x = 3/2

Therefore,

The zeroes of 2x²-3x-9 are -3 and 3/2.

$Now, sum \:of \:the \: zeroes\\=-3+\frac{3}{2}\\=\frac{-6+3}{2}\\=\frac{-3}{2}\\=\frac{-(Coefficient\:of \:x)}{coefficient\:of \:x^{2}}$

$product\:of \:the \: zeroes\\=-3\times \frac{3}{2}\\=\frac{-9}{2}\\=\frac{constant\:term}{coefficient\:of \:x^{2}}$

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