Mathematics, 07.07.2019 02:30, starreee

Which shows the correct substitution of the values a, b, and c from the equation 0 = – 3x2 – 2x + 6 into the quadratic formula? quadratic formula: x =

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$x_{1,2}=\frac{-(-2)\pm\sqrt{(-2)^2-4\cdot (-3)\cdot 6} }{2(-3)}$ shows correct substitution of the values a, b, and c from the given quadratic equation -3x^2-2x+6=0 into quadratic formula.

Step-by-step explanation:

Given: The quadratic equation -3x^2-2x+6=0

We have to show the correct substitution of the values a, b, and c from the given quadratic equation -3x^2-2x+6=0 into quadratic formula.

The standard form of quadratic equation is ax^2+bx+c=0 then the solution of quadratic equation using quadratic formula is given as $x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac} }{2a}$

Consider the given quadratic equation -3x^2-2x+6=0

Comparing with general quadratic equation, we have

a = -3 , b = -2 , c = 6

Substitute in quadratic formula, we get,

$x_{1,2}=\frac{-(-2)\pm\sqrt{(-2)^2-4\cdot (-3)\cdot 6} }{2(-3)}$

Simplify, we have,

$x_{1,2}=\frac{2\pm\sqrt{76} }{-6}$

Thus, $x_{1}=\frac{2+\sqrt{76} }{-6}$ and x_{2}=\frac{2-\sqrt{76} }{-6}

Simplify, we get,

$x_1=-\frac{1+\sqrt{19}}{3},\:x_2=\frac{\sqrt{19}-1}{3}$

Thus, $x_{1,2}=\frac{-(-2)\pm\sqrt{(-2)^2-4\cdot (-3)\cdot 6} }{2(-3)}$ shows correct substitution of the values a, b, and c from the given quadratic equation -3x^2-2x+6=0 into quadratic formula.

answered: ShelleyPSeliger

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 $x = \dfrac{-(-2) \pm \sqrt{(-2)^2 - 4 \times -3 \times 6}}{2 \times (-3)}$

What is the standard form of quadratic equations?

The standard form of quadratic equations is ax^2 + bx + c = 0

What are the solutions of standard quadratic equation?

$x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

From given equation 0 = -3x^2 -2x + 6 , we deduce that a = -3, b = -2 and c = 6.

Thus, substituting these values in the solutions' equation:

$x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\\
\\
x = \dfrac{2 \pm \sqrt{4 + 72}}{-6} = \dfrac{2\pm 2\sqrt{19}}{-6} = \dfrac{-1 \mp 2\sqrt{19}}{3}$

Thus, the correct substitution of the value a,b and c from the given quadratic equations into the solution form is $x = \dfrac{-(-2) \pm \sqrt{(-2)^2 - 4 \times -3 \times 6}}{2 \times (-3)}$

Learn more about solutions of quadratic equations here:

link

answered: Guest

Yeah me i need badly

answered: Guest

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

=3x+1

=3(2)+1

since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

=x+1

=2+1

now, let's find g(2)g(2):

g(x)

g(2)

=2x

=2⋅2

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!

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Which shows the correct substitution of the values a, b, and c from the equation 0 = – 3x2 – 2x + 6...