Answer:
[tex]x_{1,2}=\frac{-(-2)\pm\sqrt{(-2)^2-4\cdot (-3)\cdot 6} }{2(-3)}[/tex] shows correct substitution of the values a, b, and c from the given quadratic equation [tex]-3x^2-2x+6=0[/tex] into quadratic formula.
Step-by-step explanation:
Given: The quadratic equation [tex]-3x^2-2x+6=0[/tex]
We have to show the correct substitution of the values a, b, and c from the given quadratic equation [tex]-3x^2-2x+6=0[/tex] into quadratic formula.
The standard form of quadratic equation is [tex]ax^2+bx+c=0[/tex] then the solution of quadratic equation using quadratic formula is given as [tex]x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac} }{2a}[/tex]
Consider the given quadratic equation [tex]-3x^2-2x+6=0[/tex]
Comparing with general quadratic equation, we have
a = -3 , b = -2 , c = 6
Substitute in quadratic formula, we get,
[tex]x_{1,2}=\frac{-(-2)\pm\sqrt{(-2)^2-4\cdot (-3)\cdot 6} }{2(-3)}[/tex]
Simplify, we have,
[tex]x_{1,2}=\frac{2\pm\sqrt{76} }{-6}[/tex]
Thus, [tex]x_{1}=\frac{2+\sqrt{76} }{-6}[/tex] and x_{2}=\frac{2-\sqrt{76} }{-6}
Simplify, we get,
[tex]x_1=-\frac{1+\sqrt{19}}{3},\:x_2=\frac{\sqrt{19}-1}{3}[/tex]
Thus, [tex]x_{1,2}=\frac{-(-2)\pm\sqrt{(-2)^2-4\cdot (-3)\cdot 6} }{2(-3)}[/tex] shows correct substitution of the values a, b, and c from the given quadratic equation [tex]-3x^2-2x+6=0[/tex] into quadratic formula.
You can use the standard form of quadratic equation to find the correct substitution of the values a, b and c.
The correct substitution of the value a,b and c from the given quadratic equations into the solution form is [tex]x = \dfrac{-(-2) \pm \sqrt{(-2)^2 - 4 \times -3 \times 6}}{2 \times (-3)}[/tex]
The standard form of quadratic equations is [tex]ax^2 + bx + c = 0[/tex]
[tex]x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/tex]
From given equation [tex]0 = -3x^2 -2x + 6[/tex], we deduce that a = -3, b = -2 and c = 6.
Thus, substituting these values in the solutions' equation:
[tex]x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\\ \\ x = \dfrac{2 \pm \sqrt{4 + 72}}{-6} = \dfrac{2\pm 2\sqrt{19}}{-6} = \dfrac{-1 \mp 2\sqrt{19}}{3}[/tex]
Thus, the correct substitution of the value a,b and c from the given quadratic equations into the solution form is [tex]x = \dfrac{-(-2) \pm \sqrt{(-2)^2 - 4 \times -3 \times 6}}{2 \times (-3)}[/tex]
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