Question

find the zeroes of the quadratic polynomial and verify the relantionship between the zeroes of 2x^2-3x-9

Answer 1

Answer :-

• α = 3 and β = -3/2

Given :-

• A quadratic polynomial 2x² - 3x - 9

To Find :-

• The zeros of the given polynomial and verify their relationship.

Step By Step Explanation :-

We need to find the zeros of the quadratic polynomial 2x² - 3x - 9 .

So let's do it !!

$\bf \: Splitting \: the \: middle \: term \downarrow \\ \\ \implies\sf 2 {x}^{2} - 3x - 9 \\ \\ \implies\sf2 {x}^{2} - 6x + 3x - 9 \\ \\\implies\sf 2x(x - 3) + 3(x - 3) \\ \\ \implies\sf(2x + 3)(x - 3) \\ \\ \bf \: Zeros \downarrow \\ \\ \sf \: x - 3 = 0 \implies \bf x = 3 \\ \\ \sf2x + 3 = 0 \implies \bf x = \cfrac{ - 3}{2}$

Let us consider α => 3 and β => -3/2

Verification :-

As we know ⤵

$\bigstar \boxed{ \underline{ \mathfrak{ \pink{\alpha + \beta = \cfrac{ - b}{a} }}}}$

$\bigstar\boxed{ \underline{ \mathfrak{ \green{\alpha\beta = \cfrac{c}{a} }}}}$

Now let's verify it !!

By substituting the values

In equation 1 ⤵

$\implies\sf3 + \cfrac{ (- 3)}{2} = \cfrac{ - ( - 3)}{2} \\ \\\implies\sf \cfrac{6 - 3}{2} = \cfrac{3}{2} \\ \\\implies\sf \cfrac{3}{2} = \cfrac{3}{2}$

In equation 2 ⤵

$\implies\sf3 \times \cfrac{ - 3}{2} = \cfrac{ - 9}{2} \\ \\\implies\sf \cfrac{ - 9}{2} = \cfrac{ - 9}{2}$

Hence relationship verified .

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