Question

solve for the roots of the equation: 15y² - 40y + 16 = 0

Answer 1

Given Equation is in the form of ax^2 + bx + c = 0.

Where a = 15, b-40, c=16.

We know that for a quadratic equation ax^2+ bx + c = 0, the solutions are:

(1)


$x = \frac{-b+ \sqrt{b^2 - 4ac} }{2a}$

$x = \frac{-(-40) + \sqrt{(-40)^2 - 4(15)(16)} }{2(15)}$

          $= \ \textgreater \ \frac{40 + \sqrt{1600-960} }{30}$

          $= \ \textgreater \ \frac{40 + \sqrt{640} }{30}$

          $= \ \textgreater \ \frac{40 + 8 \sqrt{10} }{30}$

          $= \ \textgreater \ \frac{8(5 + \sqrt{10} )}{30}$

          $= \ \textgreater \ \frac{4(5 + \sqrt{10}) }{15}$



(2)

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

$= \ \textgreater \ \frac{-(-40)- \sqrt{(-40)^2 - 4 * 15 * 16} }{2 * 15}$

$= \ \textgreater \ \frac{40 - \sqrt{1600 - 960} }{30}$

$= \ \textgreater \ \frac{40 - \sqrt{640} }{30}$

$= \ \textgreater \ \frac{40 - 8 \sqrt{10} }{30}$

$= \ \textgreater \ \frac{8(5 - \sqrt{10}) }{30}$

$= \ \textgreater \ \frac{4(5 - \sqrt{10)} }{15}$



Hope this helps!