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Equation 14x-17+3x^2=19+11x
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![Equation log_2(4(x)^4+28)=2+log[sqrt[2],sqrt[5(x)^2+1]]](/media/krcore-image-pods/64/hash/equation/a/3b/2fd2cbe4331476c13922de3697e5d.png "Equation log_2(4(x)^4+28)=2+log[sqrt[2],sqrt[5(x)^2+1]]")
Equation log_2(4(x)^4+28)=2+log[sqrt[2],sqrt[5(x)^2+1]]
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Identical expressions
- 14x- seventeen +3x^ two = nineteen +11x
- 14x minus 17 plus 3x squared equally 19 plus 11x
- 14x minus seventeen plus 3x to the power of two equally nineteen plus 11x
- 14x-17+3x2=19+11x
- 14x-17+3x²=19+11x
- 14x-17+3x to the power of 2=19+11x
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Similar expressions
- 14x+17+3x^2=19+11x
- 14x-17+3x^2=19-11x
- 14x-17-3x^2=19+11x
14x-17+3x^2=19+11x equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Move right part of the equation to
left part with negative sign.
The equation is transformed from
$$3 x^{2} + 14 x - 17 = 11 x + 19$$
to
$$\left(- 11 x - 19\right) + \left(3 x^{2} + 14 x - 17\right) = 0$$
This equation is of the form
$$a\ x^2 + b\ x + c = 0$$
A quadratic equation can be solved using the discriminant
The roots of the quadratic equation:
$$x_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$x_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where $D = b^2 - 4 a c$ is the discriminant.
Because
$$a = 3$$
$$b = 3$$
$$c = -36$$
, then
$$D = b^2 - 4\ a\ c = $$
$$3^{2} - 3 \cdot 4 \left(-36\right) = 441$$
Because D > 0, then the equation has two roots.
$$x_1 = \frac{(-b + \sqrt{D})}{2 a}$$
$$x_2 = \frac{(-b - \sqrt{D})}{2 a}$$
or
$$x_{1} = 3$$
Simplify
$$x_{2} = -4$$
Simplify
left part with negative sign.
The equation is transformed from
$$3 x^{2} + 14 x - 17 = 11 x + 19$$
to
$$\left(- 11 x - 19\right) + \left(3 x^{2} + 14 x - 17\right) = 0$$
This equation is of the form
$$a\ x^2 + b\ x + c = 0$$
A quadratic equation can be solved using the discriminant
The roots of the quadratic equation:
$$x_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$x_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where $D = b^2 - 4 a c$ is the discriminant.
Because
$$a = 3$$
$$b = 3$$
$$c = -36$$
, then
$$D = b^2 - 4\ a\ c = $$
$$3^{2} - 3 \cdot 4 \left(-36\right) = 441$$
Because D > 0, then the equation has two roots.
$$x_1 = \frac{(-b + \sqrt{D})}{2 a}$$
$$x_2 = \frac{(-b - \sqrt{D})}{2 a}$$
or
$$x_{1} = 3$$
Simplify
$$x_{2} = -4$$
Simplify
Vieta's Theorem
rewrite the equation
$$3 x^{2} + 14 x - 17 = 11 x + 19$$
of
$$a x^{2} + b x + c = 0$$
as reduced quadratic equation
$$x^{2} + \frac{b x}{a} + \frac{c}{a} = 0$$
$$x^{2} + x - 12 = 0$$
$$p x + x^{2} + q = 0$$
where
$$p = \frac{b}{a}$$
$$p = 1$$
$$q = \frac{c}{a}$$
$$q = -12$$
Vieta Formulas
$$x_{1} + x_{2} = - p$$
$$x_{1} x_{2} = q$$
$$x_{1} + x_{2} = -1$$
$$x_{1} x_{2} = -12$$
$$3 x^{2} + 14 x - 17 = 11 x + 19$$
of
$$a x^{2} + b x + c = 0$$
as reduced quadratic equation
$$x^{2} + \frac{b x}{a} + \frac{c}{a} = 0$$
$$x^{2} + x - 12 = 0$$
$$p x + x^{2} + q = 0$$
where
$$p = \frac{b}{a}$$
$$p = 1$$
$$q = \frac{c}{a}$$
$$q = -12$$
Vieta Formulas
$$x_{1} + x_{2} = - p$$
$$x_{1} x_{2} = q$$
$$x_{1} + x_{2} = -1$$
$$x_{1} x_{2} = -12$$
Sum and product of roots
[src]
sum
-4 + 3
$$\left(-4\right) + \left(3\right)$$
=
-1
$$-1$$
product
-4 * 3
$$\left(-4\right) * \left(3\right)$$
=
-12
$$-12$$
The graph