Question

Solve the equation t+4/t+3/t-4=-16/t^2-4t

Answer 1

Answer:

The given equation can be  written as $5t^3-4t^2+7t+16=0$

The factors of the equation is

$5t^3-4t^2+7t+16=(t+1)(\frac{9+i\sqrt{239}}{10})(\frac{9-i\sqrt{239}}{10})$

Step-by-step explanation:

Given equation is $t+\frac{4}{t}+\frac{3}{t}-4=-\frac{16}{t^2}-4t$

Now to solve the given equation:

$t+\frac{4}{t}+\frac{3}{t}-4=-\frac{16}{t^2}-4t$

$t+\frac{7}{t}-4=-\frac{16}{t^2}-4t$

$\frac{t^2+7-4t}{t}=\frac{-16-4t^3}{t^2}$

Multiply the above equation into $t^2$ on both sides

$\frac{t^2+7-4t}{t}\times t^2=\frac{-16-4t^3}{t^2}\times t^2$

$t(t^2+7-4t)=-16-4t^3$

$t^3+7t-4t^2=-16-4t^3$

$t^3+7t-4t^2+16+4t^3=0$

Adding the like terms

$5t^3+7t-4t^2+16=0$

Rewritting the above equation

$5t^3-4t^2+7t+16=0$

We can solve this equation by synthetic division method

-1_|   5     -4       7     16

       0     -5       9    -16

    _______________

      5      -9       16     0

Therefore t+1 is a factor of this cubic equation

We have the quadratic equation $5t^2-9t+16=0$

For quadratic equation $ax^2+bx+c=0$ the solution is

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

where a and b are coefficients of $x^2$ and x respectively

Here a=5 ,b=-9 and c=16

$t=\frac{-(-9)\pm\sqrt{(-9)^2-4(5)(16)}}{2(5)}$

$t=\frac{9\pm\sqrt{81-320}}{10}$

$t=\frac{9\pm\sqrt{-239}}{10}$

$t=\frac{9\pm\sqrt{239i^2}}{10}$ where $i^2=-1$

$t=\frac{9\pmi\sqrt{239}}{10}$

Therefore $t=\frac{9+i\sqrt{239}}{10}$ and $t=\frac{9-i\sqrt{239}}{10}$

Therefore  the factors are t+1,$\frac{9+i\sqrt{239}}{10}$ and $\frac{9-i\sqrt{239}}{10}$

$5t^3-4t^2+7t+16=(t+1)(\frac{9+i\sqrt{239}}{10})(\frac{9-i\sqrt{239}}{10})$