Mister Exam
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-5x^2-9x+12=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
This equation is of the form
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
$$a = -5$$
$$b = -9$$
$$c = 12$$
, then
Because D > 0, then the equation has two roots.
or
$$x_{1} = - \frac{\sqrt{321}}{10} - \frac{9}{10}$$
$$x_{2} = - \frac{9}{10} + \frac{\sqrt{321}}{10}$$
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
$$a = -5$$
$$b = -9$$
$$c = 12$$
, then
D = b^2 - 4 * a * c =
(-9)^2 - 4 * (-5) * (12) = 321
Because D > 0, then the equation has two roots.
x1 = (-b + sqrt(D)) / (2*a)
x2 = (-b - sqrt(D)) / (2*a)
or
$$x_{1} = - \frac{\sqrt{321}}{10} - \frac{9}{10}$$
$$x_{2} = - \frac{9}{10} + \frac{\sqrt{321}}{10}$$
Vieta's Theorem
rewrite the equation
$$\left(- 5 x^{2} - 9 x\right) + 12 = 0$$
of
$$a x^{2} + b x + c = 0$$
as reduced quadratic equation
$$x^{2} + \frac{b x}{a} + \frac{c}{a} = 0$$
$$x^{2} + \frac{9 x}{5} - \frac{12}{5} = 0$$
$$p x + q + x^{2} = 0$$
where
$$p = \frac{b}{a}$$
$$p = \frac{9}{5}$$
$$q = \frac{c}{a}$$
$$q = - \frac{12}{5}$$
Vieta Formulas
$$x_{1} + x_{2} = - p$$
$$x_{1} x_{2} = q$$
$$x_{1} + x_{2} = - \frac{9}{5}$$
$$x_{1} x_{2} = - \frac{12}{5}$$
$$\left(- 5 x^{2} - 9 x\right) + 12 = 0$$
of
$$a x^{2} + b x + c = 0$$
as reduced quadratic equation
$$x^{2} + \frac{b x}{a} + \frac{c}{a} = 0$$
$$x^{2} + \frac{9 x}{5} - \frac{12}{5} = 0$$
$$p x + q + x^{2} = 0$$
where
$$p = \frac{b}{a}$$
$$p = \frac{9}{5}$$
$$q = \frac{c}{a}$$
$$q = - \frac{12}{5}$$
Vieta Formulas
$$x_{1} + x_{2} = - p$$
$$x_{1} x_{2} = q$$
$$x_{1} + x_{2} = - \frac{9}{5}$$
$$x_{1} x_{2} = - \frac{12}{5}$$
Rapid solution
[src]
_____
9 \/ 321
x1 = - -- + -------
10 10
$$x_{1} = - \frac{9}{10} + \frac{\sqrt{321}}{10}$$
_____
9 \/ 321
x2 = - -- - -------
10 10
$$x_{2} = - \frac{\sqrt{321}}{10} - \frac{9}{10}$$
x2 = -sqrt(321)/10 - 9/10
Sum and product of roots
[src]
sum
_____ _____ 9 \/ 321 9 \/ 321 - -- + ------- + - -- - ------- 10 10 10 10
$$\left(- \frac{\sqrt{321}}{10} - \frac{9}{10}\right) + \left(- \frac{9}{10} + \frac{\sqrt{321}}{10}\right)$$
=
-9/5
$$- \frac{9}{5}$$
product
/ _____\ / _____\ | 9 \/ 321 | | 9 \/ 321 | |- -- + -------|*|- -- - -------| \ 10 10 / \ 10 10 /
$$\left(- \frac{9}{10} + \frac{\sqrt{321}}{10}\right) \left(- \frac{\sqrt{321}}{10} - \frac{9}{10}\right)$$
=
-12/5
$$- \frac{12}{5}$$
-12/5