Question

Zeroes of the polynomial p(x) = 2 x^2 ─ 9 ─ 3x are : *

1 point

(a) 3 , 3/2

(b) ─ 3/2 , 3

(c)2 , 3

(d) 9 , 3/2

​

Answer 1

GIVEN :-

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

TO FIND :-

• The zeroes of quadratic polynomial.

SOLUTION :-

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

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

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

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

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

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

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

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

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

Answer 2

$\sf \underline{Given} :$

• Zeroes of the polynomial p(x) = 2 x^2 ─ 9 ─ 3x are : *

$\sf \underline{solution} :$

$\sf \leadsto \: 2 {x}^{2} - 9 - 3x = 0 \\ \\ \sf \leadsto \: {2x}^{2} - 3x - 9 = 0 \\ \\ \sf \leadsto \: {2x}^{2} - 6x + 3x - 9 = 0 \\ \\ \sf \leadsto \: 2x(x - 3) + 3(x - 3) = 0 \\ \\ \sf \leadsto \: (2x + 3)(x - 3) = 0 \\ \\ \sf \leadsto \: x = \frac{ - 3}{2} \: and \: x = 3$

Option b is correct - 3/2 , 3

$\rule{180}{1.5}$

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