Mister Exam
Other calculators
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How to use it?
- Equation:
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Equation cos(x)-1=0
-
Equation (x+3)^4-4(x+3)^2-5=0
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Equation x^2+14*x+24=0
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Equation (x+3)^4+2(x+3)^2-8=0
- Express {x} in terms of y where:
- -16*x-19*y=-4
- -9*x+17*y=-3
- 2*x-20*y=-7
- -1*x+1*y=18
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Identical expressions
- 5x-4x^ two + seven = one
- 5x minus 4x squared plus 7 equally 1
- 5x minus 4x to the power of two plus seven equally one
- 5x-4x2+7=1
- 5x-4x²+7=1
- 5x-4x to the power of 2+7=1
-
Similar expressions
- 5x+4x^2+7=1
- 5x-4x^2-7=1
5x-4x^2+7=1 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
$$\left(- 4 x^{2} + 5 x\right) + 7 = 1$$
to
$$\left(\left(- 4 x^{2} + 5 x\right) + 7\right) - 1 = 0$$
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
$$a = -4$$
$$b = 5$$
$$c = 6$$
, then
Because D > 0, then the equation has two roots.
or
$$x_{1} = - \frac{3}{4}$$
$$x_{2} = 2$$
left part with negative sign.
The equation is transformed from
$$\left(- 4 x^{2} + 5 x\right) + 7 = 1$$
to
$$\left(\left(- 4 x^{2} + 5 x\right) + 7\right) - 1 = 0$$
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
$$a = -4$$
$$b = 5$$
$$c = 6$$
, then
D = b^2 - 4 * a * c =
(5)^2 - 4 * (-4) * (6) = 121
Because D > 0, then the equation has two roots.
x1 = (-b + sqrt(D)) / (2*a)
x2 = (-b - sqrt(D)) / (2*a)
or
$$x_{1} = - \frac{3}{4}$$
$$x_{2} = 2$$
Vieta's Theorem
rewrite the equation
$$\left(- 4 x^{2} + 5 x\right) + 7 = 1$$
of
$$a x^{2} + b x + c = 0$$
as reduced quadratic equation
$$x^{2} + \frac{b x}{a} + \frac{c}{a} = 0$$
$$x^{2} - \frac{5 x}{4} - \frac{3}{2} = 0$$
$$p x + q + x^{2} = 0$$
where
$$p = \frac{b}{a}$$
$$p = - \frac{5}{4}$$
$$q = \frac{c}{a}$$
$$q = - \frac{3}{2}$$
Vieta Formulas
$$x_{1} + x_{2} = - p$$
$$x_{1} x_{2} = q$$
$$x_{1} + x_{2} = \frac{5}{4}$$
$$x_{1} x_{2} = - \frac{3}{2}$$
$$\left(- 4 x^{2} + 5 x\right) + 7 = 1$$
of
$$a x^{2} + b x + c = 0$$
as reduced quadratic equation
$$x^{2} + \frac{b x}{a} + \frac{c}{a} = 0$$
$$x^{2} - \frac{5 x}{4} - \frac{3}{2} = 0$$
$$p x + q + x^{2} = 0$$
where
$$p = \frac{b}{a}$$
$$p = - \frac{5}{4}$$
$$q = \frac{c}{a}$$
$$q = - \frac{3}{2}$$
Vieta Formulas
$$x_{1} + x_{2} = - p$$
$$x_{1} x_{2} = q$$
$$x_{1} + x_{2} = \frac{5}{4}$$
$$x_{1} x_{2} = - \frac{3}{2}$$
Sum and product of roots
[src]
sum
2 - 3/4
$$- \frac{3}{4} + 2$$
=
5/4
$$\frac{5}{4}$$
product
2*(-3) ------ 4
$$\frac{\left(-3\right) 2}{4}$$
=
-3/2
$$- \frac{3}{2}$$
-3/2