Mister Exam
Other calculators
-
How to use it?
- Equation:
-
Equation (x-3)*(x+4)=0
-
Equation sin(x)-1/2=0
- Equation x^2-22x-23=0
-
Equation sin5x-√3cos5x=0
- Express {x} in terms of y where:
- 20*x+6*y=8
- -1*x-17*y=-12
- -16*x-20*y=6
- 10*x+18*y=6
- Integral of d{x}:
- sqrt(x^2+8)
- Graphing y =:
- 2x+1
- Derivative of:
- 2x+1
- Canonical form:
- 2x+1
-
Identical expressions
- sqrt(x^ two + eight)=2x+ one
- square root of (x squared plus 8) equally 2x plus 1
- square root of (x to the power of two plus eight) equally 2x plus one
- √(x^2+8)=2x+1
- sqrt(x2+8)=2x+1
- sqrtx2+8=2x+1
- sqrt(x²+8)=2x+1
- sqrt(x to the power of 2+8)=2x+1
- sqrtx^2+8=2x+1
-
Similar expressions
- (76/5)*(6/5)-(1147/20)*(72/25)-(13579/50)*(457/50)-(22639/100)*(1439/100)-(4987/20)*(337/20)-1/2*x*((12913/100)+((12913/100)+(473/50)*x))*(1/2*x+(181/10))+166*(149/10)+(8352/25)*(337/20)+1/2*x*((15013/100)+((15013/100)+(1271/50)*x))*(1/2*x+(181/10))=0
- sqrt(x^2+8)=2x-1
- sqrt(x^2-8)=2x+1
sqrt(x^2+8)=2x+1 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation
$$\sqrt{x^{2} + 8} = 2 x + 1$$
$$\sqrt{x^{2} + 8} = 2 x + 1$$
We raise the equation sides to 2-th degree
$$x^{2} + 8 = \left(2 x + 1\right)^{2}$$
$$x^{2} + 8 = 4 x^{2} + 4 x + 1$$
Transfer the right side of the equation left part with negative sign
$$- 3 x^{2} - 4 x + 7 = 0$$
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
$$a = -3$$
$$b = -4$$
$$c = 7$$
, then
Because D > 0, then the equation has two roots.
or
$$x_{1} = - \frac{7}{3}$$
$$x_{2} = 1$$
Because
$$\sqrt{x^{2} + 8} = 2 x + 1$$
and
$$\sqrt{x^{2} + 8} \geq 0$$
then
$$2 x + 1 \geq 0$$
or
$$- \frac{1}{2} \leq x$$
$$x < \infty$$
The final answer:
$$x_{2} = 1$$
$$\sqrt{x^{2} + 8} = 2 x + 1$$
$$\sqrt{x^{2} + 8} = 2 x + 1$$
We raise the equation sides to 2-th degree
$$x^{2} + 8 = \left(2 x + 1\right)^{2}$$
$$x^{2} + 8 = 4 x^{2} + 4 x + 1$$
Transfer the right side of the equation left part with negative sign
$$- 3 x^{2} - 4 x + 7 = 0$$
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
$$a = -3$$
$$b = -4$$
$$c = 7$$
, then
D = b^2 - 4 * a * c =
(-4)^2 - 4 * (-3) * (7) = 100
Because D > 0, then the equation has two roots.
x1 = (-b + sqrt(D)) / (2*a)
x2 = (-b - sqrt(D)) / (2*a)
or
$$x_{1} = - \frac{7}{3}$$
$$x_{2} = 1$$
Because
$$\sqrt{x^{2} + 8} = 2 x + 1$$
and
$$\sqrt{x^{2} + 8} \geq 0$$
then
$$2 x + 1 \geq 0$$
or
$$- \frac{1}{2} \leq x$$
$$x < \infty$$
The final answer:
$$x_{2} = 1$$