The given quadratic equation is:
[tex]5x^2-7x\text{ + 1}[/tex]We can solve the equation by equating it to zero and using the quadratic formula method method
[tex]\begin{gathered} 5x^2-7x+1\text{ = 0} \\ \text{x = }\frac{\text{-b}\pm\sqrt[]{b^2-4ac}}{2a} \\ \text{x = }\frac{\text{-(-7)}\pm\sqrt[]{(-7)^2_{}^{}-4(5)(1)}}{2(5)} \\ \text{x = }\frac{7\pm\sqrt[]{49^{}_{}-20}}{10} \\ \text{x = }\frac{7\pm\sqrt[]{29}}{10} \\ \text{x = }\frac{7\pm5.385}{10} \\ x_1\text{ = }\frac{7+5.385}{10}=\frac{12.385}{10}=\text{ 1.2385} \\ x_2\text{ = }\frac{7-5.385}{10}=0.1615 \end{gathered}[/tex]The roots of the equation are:
p = 1.2385
q = 0.1615
[tex]\begin{gathered} p^2q+q^2p\text{ = (1.2385})^2(0.1615)\text{ + }(0.1615)^2(1.2385) \\ p^2q+q^2p\text{ =}0.2477\text{ + }0.0323 \\ p^2q+q^2p\text{ = }0.28 \end{gathered}[/tex]
