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- Equation:
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Identical expressions
- (a- three)²- four (a- three)+ four
- (a minus 3)² minus 4(a minus 3) plus 4
- (a minus three)² minus four (a minus three) plus four
- a-3²-4a-3+4
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Similar expressions
- (a+3)²-4(a-3)+4
- (a-3)²-4(a+3)+4
- (a-3)²-4(a-3)-4
- (a-3)²+4(a-3)+4
(a-3)²-4(a-3)+4 equation
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The solution
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2 (a - 3) - 4*(a - 3) + 4 = 0
$$\left(\left(a - 3\right)^{2} - 4 \left(a - 3\right)\right) + 4 = 0$$
Detail solution
Expand the expression in the equation
$$\left(\left(a - 3\right)^{2} - 4 \left(a - 3\right)\right) + 4 = 0$$
We get the quadratic equation
$$a^{2} - 10 a + 25 = 0$$
This equation is of the form
A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$a_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$a_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = 1$$
$$b = -10$$
$$c = 25$$
, then
Because D = 0, then the equation has one root.
$$a_{1} = 5$$
$$\left(\left(a - 3\right)^{2} - 4 \left(a - 3\right)\right) + 4 = 0$$
We get the quadratic equation
$$a^{2} - 10 a + 25 = 0$$
This equation is of the form
a*a^2 + b*a + c = 0
A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$a_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$a_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = 1$$
$$b = -10$$
$$c = 25$$
, then
D = b^2 - 4 * a * c =
(-10)^2 - 4 * (1) * (25) = 0
Because D = 0, then the equation has one root.
a = -b/2a = --10/2/(1)
$$a_{1} = 5$$