An algebra problem by renzo gantala

By Vieta's, $a + b = \dfrac{3}{4}$ and $ab = -\dfrac{7}{4}$ . Then

$\dfrac{1}{a^{2}} + \dfrac{1}{b^{2}} = \dfrac{a^{2} + b^{2}}{(ab)^{2}} = \dfrac{(a + b)^{2} - 2ab}{(ab)^{2}} = \dfrac{\left(\dfrac{3}{4}\right)^{2} - 2 \times \left(-\dfrac{7}{4}\right)}{\left(\dfrac{7}{4}\right)^{2}} = \dfrac{\dfrac{9}{16} + \dfrac{7}{2}}{\dfrac{49}{16}} = \dfrac{\dfrac{9}{16} + \dfrac{56}{16}}{\dfrac{49}{16}} = \boxed{\dfrac{65}{49}}$ .

In the equation 4x^2 - 3x - 7=0, it can be factored as (4x-7)(x+1)=0.

solving for the roots:

          4x-7=0, then the root is 7/4

          x+1=0, then the other root is -1

The reciprocals of the roots are -1 and 4/7.

To find the sum of their squares:

                                  (-1)^2+(4/7)^2=1+16/49
                                   49/49+16/49=65/49