Answer:
1. d = 40; 2 irrational
Step-by-step explanation:
[tex]\boxed{\begin{minipage}{7.6 cm}\underline{Discriminant}\\\\$b^2-4ac$ \quad when $ax^2+bx+c=0$\\\\when $b^2-4ac > 0 \implies$ two real solutions.\\when $b^2-4ac=0 \implies$ one real solution.\\when $b^2-4ac < 0 \implies$ no real solutions.\\\end{minipage}}[/tex]
Given quadratic equation:
[tex]3x^2+4x-2=0[/tex]
Therefore:
Substitute the values of a, b and c into the discriminant formula:
[tex]\begin{aligned}\implies b^2-4ac&=(4)^2-4(3)(-2)\\&=16-12(-2)\\&=16+24\\&=40\end{aligned}[/tex]
Therefore, as d = 40 and 40 > 0:
[tex]\implies b^2 - 4ac > 0 \implies \text{two real solutions}.[/tex]
To determine if the solutions are irrational or rational, simply square root the discriminant.
[tex]\implies \sqrt{d}=\sqrt{40}=2 \sqrt{10}[/tex]
As the discriminant is not a perfect square, then its square root is irrational and so the solutions of the quadratic equation are irrational.
