Question

Factorize 21x^2 - 16-7x / 3x^2-4 = 5

Answer 1

Answer: $\bold{(12x - 7 - i\sqrt{47})(12x - 7 + i\sqrt{47})=0}$

Step-by-step explanation:

$\dfrac{21x^2-7x - 16}{3x^2-4}=5$

cross multiply:  21x² - 7x - 16 = 15x² - 20

set equal to 0:  6x² - 7x + 4 = 0

Use quadratic formula to find the roots:

a=6, b=-7, c=4

$x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$

   $=\dfrac{-(-7) \pm \sqrt{(-7)^2-4(6)(4)}}{2(6)}$

   $=\dfrac{7 \pm \sqrt{49-96}}{12}$

   $=\dfrac{7 \pm \sqrt{-47}}{12}$

   $=\dfrac{7 \pm i\sqrt{47}}{12}$

$x_1 =\dfrac{7 + i\sqrt{47}}{12} \qquad >>\qquad (12x - 7 - i\sqrt{47})= 0$

$x_2 =\dfrac{7 - i\sqrt{47}}{12} \qquad >>\qquad (12x - 7 + i\sqrt{47})= 0$