Question

zeroes of the polynomial p(x)=2x^2-9-3x are​

Answer 1

Answer:

2X^2-3x - 9

2x^2 _6x + 3x - 9

2x ( x _3) + 3( x - 3 )

(x-3) (2x + 3 ) .

X= 3 , x= -3 / 2

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Answer 2

Step-by-step explanation:

GIVEN :-

• A quadratic polynomial = 2x² - 9 - 3x.

TO FIND :-

• The zeroes of quadratic polynomial.

SOLUTION :-

$\\ : \implies \displaystyle \sf \: 2x ^{2} - 9 - 3x = 0$

$: \implies \displaystyle \sf \: 2x ^{2} - 3x - 9 = 0$

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

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

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

$: \implies \displaystyle \sf \: 2x + 3 = 0 \: , \: x - 3 = 0$

$: \implies \displaystyle \sf \:2x = 0 - 3 \: , \: x = 0 + 3$

$: \implies \displaystyle \sf \:2x = - 3 \: , \: x = 3$

$: \implies \underline{ \boxed{\displaystyle \sf \bold{\:x = \frac{ - 3}{2} \: , \: x = 3}}}$

$\boxed{\begin{minipage}{5.5 cm} { \bigstar\: \textsf{For a Quadratic Polynomial :}}\\\\ {\qquad\sf p(x) = ax ^\sf2 \sf + bx + c}\\\sf with zeroes \alpha\:\sf and\:\beta \\\\\\ {\textcircled{\footnotesize1}} \:\:\alpha +\beta= \dfrac{ - \:b}{a}\:\:\bigg\lgroup\bf Sum\:of\:Zeroes\bigg\rgroup \\\\\\{\textcircled{\footnotesize2}} \: \:\alpha \beta= \sf\dfrac{c}{a}\:\:\bigg\lgroup\bf Product\:of\:Zeroes\bigg\rgroup\end{minipage}}$