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- Equation:
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Equation x^2-x-7=0
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Equation x^2+x-9=0
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Equation -2(-2-y)-y-2=0
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Equation x+5=13
- Express {x} in terms of y where:
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- -7*x+5*y=1
- -6*x-19*y=7
- -19*x-4*y=-19
- Integral of d{x}:
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3x
- Graphing y =:
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- Derivative of:
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3x
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Identical expressions
- (4x- three)(4x+ three)-(4x- one)=3x
- (4x minus 3)(4x plus 3) minus (4x minus 1) equally 3x
- (4x minus three)(4x plus three) minus (4x minus one) equally 3x
- 4x-34x+3-4x-1=3x
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Similar expressions
- 3(x+1)(x+2)=2x^2+4x
- (4x-3)(4x+3)+(4x-1)=3x
- (4x-3)(4x+3)-(4x+1)=3x
- (2*x+3)/(x+2)=(3*x+2)/x
- (4x-3)(4x-3)-(4x-1)=3x
- (4x+3)(4x+3)-(4x-1)=3x
(4x-3)(4x+3)-(4x-1)=3x equation
The teacher will be very surprised to see your correct solution 😉
The solution
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[src]
(4*x - 3)*(4*x + 3) + -4*x + 1 = 3*x
$$\left(1 - 4 x\right) + \left(4 x - 3\right) \left(4 x + 3\right) = 3 x$$
Detail solution
Move right part of the equation to
left part with negative sign.
The equation is transformed from
$$\left(1 - 4 x\right) + \left(4 x - 3\right) \left(4 x + 3\right) = 3 x$$
to
$$- 3 x + \left(\left(1 - 4 x\right) + \left(4 x - 3\right) \left(4 x + 3\right)\right) = 0$$
Expand the expression in the equation
$$- 3 x + \left(\left(1 - 4 x\right) + \left(4 x - 3\right) \left(4 x + 3\right)\right) = 0$$
We get the quadratic equation
$$16 x^{2} - 7 x - 8 = 0$$
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
$$a = 16$$
$$b = -7$$
$$c = -8$$
, then
Because D > 0, then the equation has two roots.
or
$$x_{1} = \frac{7}{32} + \frac{\sqrt{561}}{32}$$
$$x_{2} = \frac{7}{32} - \frac{\sqrt{561}}{32}$$
left part with negative sign.
The equation is transformed from
$$\left(1 - 4 x\right) + \left(4 x - 3\right) \left(4 x + 3\right) = 3 x$$
to
$$- 3 x + \left(\left(1 - 4 x\right) + \left(4 x - 3\right) \left(4 x + 3\right)\right) = 0$$
Expand the expression in the equation
$$- 3 x + \left(\left(1 - 4 x\right) + \left(4 x - 3\right) \left(4 x + 3\right)\right) = 0$$
We get the quadratic equation
$$16 x^{2} - 7 x - 8 = 0$$
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
$$a = 16$$
$$b = -7$$
$$c = -8$$
, then
D = b^2 - 4 * a * c =
(-7)^2 - 4 * (16) * (-8) = 561
Because D > 0, then the equation has two roots.
x1 = (-b + sqrt(D)) / (2*a)
x2 = (-b - sqrt(D)) / (2*a)
or
$$x_{1} = \frac{7}{32} + \frac{\sqrt{561}}{32}$$
$$x_{2} = \frac{7}{32} - \frac{\sqrt{561}}{32}$$
Rapid solution
[src]
_____
7 \/ 561
x1 = -- - -------
32 32
$$x_{1} = \frac{7}{32} - \frac{\sqrt{561}}{32}$$
_____
7 \/ 561
x2 = -- + -------
32 32
$$x_{2} = \frac{7}{32} + \frac{\sqrt{561}}{32}$$
x2 = 7/32 + sqrt(561)/32
Sum and product of roots
[src]
sum
_____ _____ 7 \/ 561 7 \/ 561 -- - ------- + -- + ------- 32 32 32 32
$$\left(\frac{7}{32} - \frac{\sqrt{561}}{32}\right) + \left(\frac{7}{32} + \frac{\sqrt{561}}{32}\right)$$
=
7/16
$$\frac{7}{16}$$
product
/ _____\ / _____\ |7 \/ 561 | |7 \/ 561 | |-- - -------|*|-- + -------| \32 32 / \32 32 /
$$\left(\frac{7}{32} - \frac{\sqrt{561}}{32}\right) \left(\frac{7}{32} + \frac{\sqrt{561}}{32}\right)$$
=
-1/2
$$- \frac{1}{2}$$
-1/2