Question

find the zeros of quadratic polynomial $3x {?}^{2} { - 5x} - 1{2}$​

Answer 1

Answer:

First write the given equation correctly

$\qquad\quad {:}\dashrightarrow\sf 3x^2-5x-12=0$

Now Factorise

$\qquad\quad {:}\dashrightarrow\sf 3x^2+9x-4x-12=0$

$\qquad\quad {:}\dashrightarrow\sf 3x (x+3)-4 (x+3)=0$

$\qquad\quad {:}\dashrightarrow\sf (x+3)(3x-4)=0$

$\qquad\quad {:}\dashrightarrow\sf (x+3)=0\quad or\quad (3x-4)=0$

$\qquad\quad {:}\dashrightarrow\sf x=-3\quad or\quad=x={\dfrac{4}{3}}$

$\therefore{\underline{\boxed{\sf x=-3\:or\;{\dfrac {4}{3}}}}}$

Answer 2

$\huge\bf{{\color{navy}{a}}{\color{indigo}{n}}{\color{blue}{s}}{\blue{w}}{\color{skyblue}{e}}{\color{lightblue}{r}}}$

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$\large\bf{To~find:}$

The zeros of the quadratic polynomial :

$\bf 3x^{2} - 5x - 12=0$

By factorising it by ‛Splitting the middle term method’ we get :

$\bf s = - 5 \: \: \: \: \: \: \: p = - 36 \\ \bf ( - 9 \: \: \: 4) \\ \bf (3x^{2} - 9x) + (4x - 12) = 0\\ \bf [3x(x - 3)] + [4(x - 3)] = 0 \\ \bf (3x + 4)(x - 3) = 0 \\ \bf 3x + 4 = 0 \: \: \: \: \: \: or \: \: \: \: \: \: x - 3 = 0 \\ \\ \bf x = \frac{ - 4}{3} \: \: \: \: \: \: \: or \: \: \: \: \: \: \: x = 3$

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$\huge\bf{{\color{navy}{Hope}}~{\color{blue}{it}}~{\blue{helps..!!}}}$