Mister Exam

7x^2+6x-9=-x^2+14-3 equation

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Numerical solution:

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The solution

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   2                2
7*x  + 6*x - 9 = - x  + 14 - 3
$$\left(7 x^{2} + 6 x\right) - 9 = \left(14 - x^{2}\right) - 3$$
Detail solution
Move right part of the equation to
left part with negative sign.

The equation is transformed from
$$\left(7 x^{2} + 6 x\right) - 9 = \left(14 - x^{2}\right) - 3$$
to
$$\left(\left(x^{2} - 14\right) + 3\right) + \left(\left(7 x^{2} + 6 x\right) - 9\right) = 0$$
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 - it is the discriminant.
Because
$$a = 8$$
$$b = 6$$
$$c = -20$$
, then
D = b^2 - 4 * a * c =

(6)^2 - 4 * (8) * (-20) = 676

Because D > 0, then the equation has two roots.
x1 = (-b + sqrt(D)) / (2*a)

x2 = (-b - sqrt(D)) / (2*a)

or
$$x_{1} = \frac{5}{4}$$
$$x_{2} = -2$$
Vieta's Theorem
rewrite the equation
$$\left(7 x^{2} + 6 x\right) - 9 = \left(14 - x^{2}\right) - 3$$
of
$$a x^{2} + b x + c = 0$$
$$x^{2} + \frac{b x}{a} + \frac{c}{a} = 0$$
$$x^{2} + \frac{3 x}{4} - \frac{5}{2} = 0$$
$$p x + q + x^{2} = 0$$
where
$$p = \frac{b}{a}$$
$$p = \frac{3}{4}$$
$$q = \frac{c}{a}$$
$$q = - \frac{5}{2}$$
Vieta Formulas
$$x_{1} + x_{2} = - p$$
$$x_{1} x_{2} = q$$
$$x_{1} + x_{2} = - \frac{3}{4}$$
$$x_{1} x_{2} = - \frac{5}{2}$$
The graph
Rapid solution [src]
x1 = -2
$$x_{1} = -2$$
x2 = 5/4
$$x_{2} = \frac{5}{4}$$
x2 = 5/4
Sum and product of roots [src]
sum
-2 + 5/4
$$-2 + \frac{5}{4}$$
=
-3/4
$$- \frac{3}{4}$$
product
-2*5
----
4  
$$- \frac{5}{2}$$
=
-5/2
$$- \frac{5}{2}$$
-5/2
x1 = -2.0
x2 = 1.25
x2 = 1.25