Answer:
The quadratic equation 3x² - 12x + 9 = 0 has two real roots when solved:
x₁ = 1 and x₂ = 3
Step-by-step explanation:
✍ An equation of type ax² + bx + c = 0, can be solved, for example, using the quadratic formula:
[tex]\boldsymbol{x=\dfrac{-b\pm\sqrt{b^{2}-4ac } }{2a} }[/tex]
either
[tex]\boldsymbol{x=\dfrac{-b\pm\sqrt{\Delta} }{2a} }[/tex]
where
[tex]\bf{\Delta=b^{2}-4ac }[/tex]
Identify the coefficients
a = 3, b = -12 and c = 9
Calculate the discriminant value
Δ = b² - 4ac
Δ = (-12)² - 4.3.9 = 144 - 12.9
Δ = 144 - 108 = 36
Enter the values of a, b and the discriminant value in the quadratic formula
[tex]\boldsymbol{x=\dfrac{-b\pm\sqrt{\Delta } }{2a} }[/tex]
[tex]\boldsymbol{x=\dfrac{-(-12\pm\sqrt{36} }{2\cdot3 } }[/tex]
[tex]\boldsymbol{x=\dfrac{(12\pm\sqrt{36} }{6 } \ ==== > \ (Solution \ general) }[/tex]
As we can see above, the discriminant (Δ) of this equation is positive (Δ> 0) which means that there are two real roots (two solutions), x₁ and x₂.
[tex]\boldsymbol{x_{1}=\dfrac{-12-\sqrt{36} }{6}=\dfrac{12-6}{6}=\dfrac{6}{6}=1 \ === > (First \ solution) }[/tex]
To find x₁, just choose the positive sign before the square root. Later,
[tex]\boldsymbol{x_{1}=\dfrac{12+\sqrt{36} }{6}=\dfrac{12+6}{6}=\dfrac{18}{6}=3 \ === > (Second \ solution) }[/tex]
