Mister Exam
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How to use it?
- Equation:
- Equation x-y=7
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Equation 1/(x+6)=2
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Equation (x-4)^2-x^2=0
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Equation 6/(x+5)=-5
- Express {x} in terms of y where:
- 7*x-1*y=-7
- 10*x-15*y=16
- -14*x-12*y=5
- 13*x-12*y=19
- Derivative of:
- 2x+1
- Graphing y =:
- 2x+1
- Canonical form:
- 2x+1
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Identical expressions
- (x^ two + three x+3)^(zero . five)=2x+ one
- (x squared plus 3x plus 3) to the power of (0.5) equally 2x plus 1
- (x to the power of two plus three x plus 3) to the power of (zero . five) equally 2x plus one
- (x2+3x+3)(0.5)=2x+1
- x2+3x+30.5=2x+1
- (x²+3x+3)^(0.5)=2x+1
- (x to the power of 2+3x+3) to the power of (0.5)=2x+1
- x^2+3x+3^0.5=2x+1
-
Similar expressions
- (x^2-3x+3)^(0.5)=2x+1
- (x^2+3x+3)^(0.5)=2x-1
- (x^2+3x-3)^(0.5)=2x+1
- 2x+5=2(x+1)+11
- 8^x-9*2^(x+1)+2^(5-x)=0
(x^2+3x+3)^(0.5)=2x+1 equation
The teacher will be very surprised to see your correct solution 😉
The solution
You have entered
[src]
______________ / 2 \/ x + 3*x + 3 = 2*x + 1
$$\sqrt{\left(x^{2} + 3 x\right) + 3} = 2 x + 1$$
Detail solution
Given the equation
$$\sqrt{\left(x^{2} + 3 x\right) + 3} = 2 x + 1$$
$$\sqrt{x^{2} + 3 x + 3} = 2 x + 1$$
We raise the equation sides to 2-th degree
$$x^{2} + 3 x + 3 = \left(2 x + 1\right)^{2}$$
$$x^{2} + 3 x + 3 = 4 x^{2} + 4 x + 1$$
Transfer the right side of the equation left part with negative sign
$$- 3 x^{2} - x + 2 = 0$$
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
$$a = -3$$
$$b = -1$$
$$c = 2$$
, then
Because D > 0, then the equation has two roots.
or
$$x_{1} = -1$$
$$x_{2} = \frac{2}{3}$$
Because
$$\sqrt{x^{2} + 3 x + 3} = 2 x + 1$$
and
$$\sqrt{x^{2} + 3 x + 3} \geq 0$$
then
$$2 x + 1 \geq 0$$
or
$$- \frac{1}{2} \leq x$$
$$x < \infty$$
The final answer:
$$x_{2} = \frac{2}{3}$$
$$\sqrt{\left(x^{2} + 3 x\right) + 3} = 2 x + 1$$
$$\sqrt{x^{2} + 3 x + 3} = 2 x + 1$$
We raise the equation sides to 2-th degree
$$x^{2} + 3 x + 3 = \left(2 x + 1\right)^{2}$$
$$x^{2} + 3 x + 3 = 4 x^{2} + 4 x + 1$$
Transfer the right side of the equation left part with negative sign
$$- 3 x^{2} - x + 2 = 0$$
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
$$a = -3$$
$$b = -1$$
$$c = 2$$
, then
D = b^2 - 4 * a * c =
(-1)^2 - 4 * (-3) * (2) = 25
Because D > 0, then the equation has two roots.
x1 = (-b + sqrt(D)) / (2*a)
x2 = (-b - sqrt(D)) / (2*a)
or
$$x_{1} = -1$$
$$x_{2} = \frac{2}{3}$$
Because
$$\sqrt{x^{2} + 3 x + 3} = 2 x + 1$$
and
$$\sqrt{x^{2} + 3 x + 3} \geq 0$$
then
$$2 x + 1 \geq 0$$
or
$$- \frac{1}{2} \leq x$$
$$x < \infty$$
The final answer:
$$x_{2} = \frac{2}{3}$$
Sum and product of roots
[src]
sum
2/3
$$\frac{2}{3}$$
=
2/3
$$\frac{2}{3}$$
product
2/3
$$\frac{2}{3}$$
=
2/3
$$\frac{2}{3}$$
2/3