Find roots of polynomial $$ p(x) = 9x^2+1488x^2-2500 $$
Answer
The roots of polynomial $ p(x) $ are:
$$ \begin{aligned}x_1 &= \frac{ 50 \sqrt{ 1497}}{ 1497 }\\[1 em]x_2 &= -50 \frac{\sqrt{ 1497 }}{ 1497 } \end{aligned} $$Explanation
Step 1:
Combine like terms:
$$ \color{blue}{9x^2} + \color{blue}{1488x^2} -2500 = \color{blue}{1497x^2} -2500 $$Step 2:
The solutions of $ 1497x^2-2500 = 0 $ are: $ x = -50 \dfrac{\sqrt{ 1497 }}{ 1497 } ~ \text{and} ~ x = \dfrac{ 50 \sqrt{ 1497}}{ 1497 }$.
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