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- Equation:
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Equation 2x^2+15-3x=11x-5
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Identical expressions
- two x^2+ fifteen -3x=11x- five
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- two x squared plus fifteen minus 3x equally 11x minus five
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Similar expressions
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- 2x^2+15+3x=11x-5
2x^2+15-3x=11x-5 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
$$2 x^{2} - 3 x + 15 = 11 x - 5$$
to
$$\left(- 11 x + 5\right) + \left(2 x^{2} - 3 x + 15\right) = 0$$
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
$$a = 2$$
$$b = -14$$
$$c = 20$$
, then
$$D = b^2 - 4\ a\ c = $$
$$\left(-1\right) 2 \cdot 4 \cdot 20 + \left(-14\right)^{2} = 36$$
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
$$x_{1} = 5$$
Simplify
$$x_{2} = 2$$
Simplify
left part with negative sign.
The equation is transformed from
$$2 x^{2} - 3 x + 15 = 11 x - 5$$
to
$$\left(- 11 x + 5\right) + \left(2 x^{2} - 3 x + 15\right) = 0$$
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
$$a = 2$$
$$b = -14$$
$$c = 20$$
, then
$$D = b^2 - 4\ a\ c = $$
$$\left(-1\right) 2 \cdot 4 \cdot 20 + \left(-14\right)^{2} = 36$$
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
$$x_{1} = 5$$
Simplify
$$x_{2} = 2$$
Simplify
Vieta's Theorem
rewrite the equation
$$2 x^{2} - 3 x + 15 = 11 x - 5$$
of
$$a x^{2} + b x + c = 0$$
as reduced quadratic equation
$$x^{2} + \frac{b x}{a} + \frac{c}{a} = 0$$
$$x^{2} - 7 x + 10 = 0$$
$$p x + x^{2} + q = 0$$
where
$$p = \frac{b}{a}$$
$$p = -7$$
$$q = \frac{c}{a}$$
$$q = 10$$
Vieta Formulas
$$x_{1} + x_{2} = - p$$
$$x_{1} x_{2} = q$$
$$x_{1} + x_{2} = 7$$
$$x_{1} x_{2} = 10$$
$$2 x^{2} - 3 x + 15 = 11 x - 5$$
of
$$a x^{2} + b x + c = 0$$
as reduced quadratic equation
$$x^{2} + \frac{b x}{a} + \frac{c}{a} = 0$$
$$x^{2} - 7 x + 10 = 0$$
$$p x + x^{2} + q = 0$$
where
$$p = \frac{b}{a}$$
$$p = -7$$
$$q = \frac{c}{a}$$
$$q = 10$$
Vieta Formulas
$$x_{1} + x_{2} = - p$$
$$x_{1} x_{2} = q$$
$$x_{1} + x_{2} = 7$$
$$x_{1} x_{2} = 10$$
Sum and product of roots
[src]
sum
2 + 5
$$\left(2\right) + \left(5\right)$$
=
7
$$7$$
product
2 * 5
$$\left(2\right) * \left(5\right)$$
=
10
$$10$$
The graph