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Identical expressions
- sqrt(4x^ two -9x+ two)=x- two
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Similar expressions
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sqrt(4x^2-9x+2)=x-2 equation
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________________ / 2 \/ 4*x - 9*x + 2 = x - 2
$$\sqrt{\left(4 x^{2} - 9 x\right) + 2} = x - 2$$
Detail solution
Given the equation
$$\sqrt{\left(4 x^{2} - 9 x\right) + 2} = x - 2$$
$$\sqrt{4 x^{2} - 9 x + 2} = x - 2$$
We raise the equation sides to 2-th degree
$$4 x^{2} - 9 x + 2 = \left(x - 2\right)^{2}$$
$$4 x^{2} - 9 x + 2 = x^{2} - 4 x + 4$$
Transfer the right side of the equation left part with negative sign
$$3 x^{2} - 5 x - 2 = 0$$
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
$$a = 3$$
$$b = -5$$
$$c = -2$$
, then
Because D > 0, then the equation has two roots.
or
$$x_{1} = 2$$
$$x_{2} = - \frac{1}{3}$$
Because
$$\sqrt{4 x^{2} - 9 x + 2} = x - 2$$
and
$$\sqrt{4 x^{2} - 9 x + 2} \geq 0$$
then
$$x - 2 \geq 0$$
or
$$2 \leq x$$
$$x < \infty$$
The final answer:
$$x_{1} = 2$$
$$\sqrt{\left(4 x^{2} - 9 x\right) + 2} = x - 2$$
$$\sqrt{4 x^{2} - 9 x + 2} = x - 2$$
We raise the equation sides to 2-th degree
$$4 x^{2} - 9 x + 2 = \left(x - 2\right)^{2}$$
$$4 x^{2} - 9 x + 2 = x^{2} - 4 x + 4$$
Transfer the right side of the equation left part with negative sign
$$3 x^{2} - 5 x - 2 = 0$$
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
$$a = 3$$
$$b = -5$$
$$c = -2$$
, then
D = b^2 - 4 * a * c =
(-5)^2 - 4 * (3) * (-2) = 49
Because D > 0, then the equation has two roots.
x1 = (-b + sqrt(D)) / (2*a)
x2 = (-b - sqrt(D)) / (2*a)
or
$$x_{1} = 2$$
$$x_{2} = - \frac{1}{3}$$
Because
$$\sqrt{4 x^{2} - 9 x + 2} = x - 2$$
and
$$\sqrt{4 x^{2} - 9 x + 2} \geq 0$$
then
$$x - 2 \geq 0$$
or
$$2 \leq x$$
$$x < \infty$$
The final answer:
$$x_{1} = 2$$