Question

find the zeroes of the following quadratic polynomials and verify the relation between zeroes and coefficients. a ) 4x²+12x+9​

Answer 1

Answer

• Zeros of the quadratic polynomial are $\sf \cfrac{-3}{2} \: and \: \cfrac{-3}{2}$

Given

• A quadratic polynomial 4x² + 12x + 9.

To Do

• To find the zeros of the quadratic polynomial and to verify the relation between zeros and coefficients.

Step By Step Explanation

Zeros :

Let's find the zeros of the quadratic polynomial 4x² + 12x + 9.

So let's do it !!

$\longmapsto \sf4 {x}^{2} + 12x + 9 \\ \\ \bigstar \: \: \underline{\boxed{ \red {\bold{{(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy}}}} \\ \\ \longmapsto \sf {(2x)}^{2} + 2 \times 2x \times 3 + {(3)}^{2} \\ \\ \longmapsto\sf{(2x + 3)}^{2} \\ \\ \longmapsto\sf (2x + 3)(2x + 3) \\ \\ \bold{Zeros \: of \: the \: polynomial \: are\downarrow} \\ \\ \longmapsto \sf \purple{2x + 3 = 0 \implies \bold{ x = \cfrac{ - 3}{2}}} \\ \\ \longmapsto \sf \pink{2x + 3 = 0 \implies \bold{x = \cfrac{ - 3}{2}}}$

Verification :

On comparing with ax² + bx + c.

a = 4 , b = 12 and c = 9.

So let's verify :

$\bigstar \: \: \: \: \underline{\boxed{ \bold{ \alpha + \beta = \cfrac{ - b}{a}}}}$

By substituting the values :

$\longmapsto \sf\cfrac{ - 3}{2} + \cfrac{ - 3}{2} = \cfrac{ - 12}{4} \\ \\ \longmapsto \sf \cfrac{ - 3 - 3}{2} = \cfrac{ - 12}{4} \\ \\ \longmapsto \sf \cfrac{ - \cancel6}{ \cancel2} = \cfrac{ - \cancel{12}}{ \cancel4} \\ \\ \longmapsto \underline {\boxed{\bf{ - 3 = - 3}}} \: \: \: \: \dag$

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$\bigstar \: \: \: \: \underline{ \boxed{ \bold{ \alpha \beta = \cfrac{c}{a}}}}$

By substituting the values :

$\longmapsto \sf\cfrac{ - 3}{2} \times \cfrac{ - 3}{2} = \cfrac{9}{4} \\ \\ \longmapsto \underline{\boxed{ \bf{\cfrac{9}{4} = \cfrac{9}{4}}}} \: \: \: \: \dag$

Hence, Verified.

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