Mister Exam
(3x^2-20x-7)/(x^2-16x+63)<0 inequation
The solution
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2 3*x - 20*x - 7 --------------- < 0 2 x - 16*x + 63
$$\frac{\left(3 x^{2} - 20 x\right) - 7}{\left(x^{2} - 16 x\right) + 63} < 0$$
(3*x^2 - 20*x - 7)/(x^2 - 16*x + 63) < 0
Detail solution
Given the inequality:
$$\frac{\left(3 x^{2} - 20 x\right) - 7}{\left(x^{2} - 16 x\right) + 63} < 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\frac{\left(3 x^{2} - 20 x\right) - 7}{\left(x^{2} - 16 x\right) + 63} = 0$$
Solve:
Given the equation:
$$\frac{\left(3 x^{2} - 20 x\right) - 7}{\left(x^{2} - 16 x\right) + 63} = 0$$
Multiply the equation sides by the denominators:
63 + x^2 - 16*x
we get:
$$\frac{\left(\left(3 x^{2} - 20 x\right) - 7\right) \left(x^{2} - 16 x + 63\right)}{\left(x^{2} - 16 x\right) + 63} = 0$$
$$3 x^{2} - 20 x - 7 = 0$$
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
$$a = 3$$
$$b = -20$$
$$c = -7$$
, then
Because D > 0, then the equation has two roots.
or
$$x_{1} = 7$$
$$x_{2} = - \frac{1}{3}$$
$$x_{1} = 7$$
$$x_{2} = - \frac{1}{3}$$
$$x_{1} = 7$$
$$x_{2} = - \frac{1}{3}$$
This roots
$$x_{2} = - \frac{1}{3}$$
$$x_{1} = 7$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} < x_{2}$$
For example, let's take the point
$$x_{0} = x_{2} - \frac{1}{10}$$
=
$$- \frac{1}{3} + - \frac{1}{10}$$
=
$$- \frac{13}{30}$$
substitute to the expression
$$\frac{\left(3 x^{2} - 20 x\right) - 7}{\left(x^{2} - 16 x\right) + 63} < 0$$
$$\frac{-7 + \left(3 \left(- \frac{13}{30}\right)^{2} - \frac{\left(-13\right) 20}{30}\right)}{\left(\left(- \frac{13}{30}\right)^{2} - \frac{\left(-13\right) 16}{30}\right) + 63} < 0$$
but
Then
$$x < - \frac{1}{3}$$
no execute
one of the solutions of our inequality is:
$$x > - \frac{1}{3} \wedge x < 7$$
$$\frac{\left(3 x^{2} - 20 x\right) - 7}{\left(x^{2} - 16 x\right) + 63} < 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\frac{\left(3 x^{2} - 20 x\right) - 7}{\left(x^{2} - 16 x\right) + 63} = 0$$
Solve:
Given the equation:
$$\frac{\left(3 x^{2} - 20 x\right) - 7}{\left(x^{2} - 16 x\right) + 63} = 0$$
Multiply the equation sides by the denominators:
63 + x^2 - 16*x
we get:
$$\frac{\left(\left(3 x^{2} - 20 x\right) - 7\right) \left(x^{2} - 16 x + 63\right)}{\left(x^{2} - 16 x\right) + 63} = 0$$
$$3 x^{2} - 20 x - 7 = 0$$
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
$$a = 3$$
$$b = -20$$
$$c = -7$$
, then
D = b^2 - 4 * a * c =
(-20)^2 - 4 * (3) * (-7) = 484
Because D > 0, then the equation has two roots.
x1 = (-b + sqrt(D)) / (2*a)
x2 = (-b - sqrt(D)) / (2*a)
or
$$x_{1} = 7$$
$$x_{2} = - \frac{1}{3}$$
$$x_{1} = 7$$
$$x_{2} = - \frac{1}{3}$$
$$x_{1} = 7$$
$$x_{2} = - \frac{1}{3}$$
This roots
$$x_{2} = - \frac{1}{3}$$
$$x_{1} = 7$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} < x_{2}$$
For example, let's take the point
$$x_{0} = x_{2} - \frac{1}{10}$$
=
$$- \frac{1}{3} + - \frac{1}{10}$$
=
$$- \frac{13}{30}$$
substitute to the expression
$$\frac{\left(3 x^{2} - 20 x\right) - 7}{\left(x^{2} - 16 x\right) + 63} < 0$$
$$\frac{-7 + \left(3 \left(- \frac{13}{30}\right)^{2} - \frac{\left(-13\right) 20}{30}\right)}{\left(\left(- \frac{13}{30}\right)^{2} - \frac{\left(-13\right) 16}{30}\right) + 63} < 0$$
9/283 < 0
but
9/283 > 0
Then
$$x < - \frac{1}{3}$$
no execute
one of the solutions of our inequality is:
$$x > - \frac{1}{3} \wedge x < 7$$
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Solving inequality on a graph
Rapid solution 2
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(-1/3, 7) U (7, 9)
$$x\ in\ \left(- \frac{1}{3}, 7\right) \cup \left(7, 9\right)$$
x in Union(Interval.open(-1/3, 7), Interval.open(7, 9))
Rapid solution
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Or(And(-1/3 < x, x < 7), And(7 < x, x < 9))
$$\left(- \frac{1}{3} < x \wedge x < 7\right) \vee \left(7 < x \wedge x < 9\right)$$
((-1/3 < x)∧(x < 7))∨((7 < x)∧(x < 9))