Question
Simplify $3 x(4 x-5)+3$ and find its values for
(a) $x=3$ $\quad$ (b) $x=\frac{1}{2}$
(a) $x=3$ $\quad$ (b) $x=\frac{1}{2}$
✓
Answer
(i) We have, $3 x(4 x-5)+3$
$=(3 x) \times(4 x)-(3 x) \times(5)+3\qquad$ [by distributive law]
$=(3 \times 4) \times(x \times x)-(3 \times 5) \times(x)+3 $
$ =12 x^2-15 x+3$
(a) When $x=3$ then
$12 x^2-15 x+3 =12(3)^2-15(3)+3 $
$ =(12 \times 9)-45+3 $
$ =108-45+3=111-45=66$
(b) When $x=\frac{1}{2},$ then
$12x^{2}-15x+3=12(\frac{1}{2})^{2}-15(\frac{1}{2})+3$
$=12 \times \frac{1}{4}-\frac{15}{2}+3=3-\frac{15}{2}+3$
$=6-\frac{15}{2}=\frac{12-15}{2}=-\frac{3}{2}$
$=(3 x) \times(4 x)-(3 x) \times(5)+3\qquad$ [by distributive law]
$=(3 \times 4) \times(x \times x)-(3 \times 5) \times(x)+3 $
$ =12 x^2-15 x+3$
(a) When $x=3$ then
$12 x^2-15 x+3 =12(3)^2-15(3)+3 $
$ =(12 \times 9)-45+3 $
$ =108-45+3=111-45=66$
(b) When $x=\frac{1}{2},$ then
$12x^{2}-15x+3=12(\frac{1}{2})^{2}-15(\frac{1}{2})+3$
$=12 \times \frac{1}{4}-\frac{15}{2}+3=3-\frac{15}{2}+3$
$=6-\frac{15}{2}=\frac{12-15}{2}=-\frac{3}{2}$
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