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Given the equation, 3x2 + 4x – 2 = 0, calculate the discriminant and determine the number and nature of the solutions.

1.d = 40; 2 irrational
2.d = 40; 2 rational
3.d = −8; 2 complex

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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:

  • a = 3
  • b = 4
  • c = -2

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.

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