Mister Exam
Other calculators
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How to use it?
- Equation:
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Equation (-x-5)(2x+4)=0
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Equation k^2-2*k+2=0
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Equation k^2-4*k+3=0
- Express {x} in terms of y where:
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Identical expressions
- k^ two - four *k+ three = zero
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- k2-4*k+3=0
- k²-4*k+3=0
- k to the power of 2-4*k+3=0
- k^2-4k+3=0
- k2-4k+3=0
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Similar expressions
- k^2+4*k+3=0
- k^2-4*k-3=0
k^2-4*k+3=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
This equation is of the form
A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$k_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$k_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = 1$$
$$b = -4$$
$$c = 3$$
, then
Because D > 0, then the equation has two roots.
or
$$k_{1} = 3$$
Simplify
$$k_{2} = 1$$
Simplify
a*k^2 + b*k + c = 0
A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$k_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$k_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = 1$$
$$b = -4$$
$$c = 3$$
, then
D = b^2 - 4 * a * c =
(-4)^2 - 4 * (1) * (3) = 4
Because D > 0, then the equation has two roots.
k1 = (-b + sqrt(D)) / (2*a)
k2 = (-b - sqrt(D)) / (2*a)
or
$$k_{1} = 3$$
Simplify
$$k_{2} = 1$$
Simplify
Vieta's Theorem
it is reduced quadratic equation
$$k^{2} + k p + q = 0$$
where
$$p = \frac{b}{a}$$
$$p = -4$$
$$q = \frac{c}{a}$$
$$q = 3$$
Vieta Formulas
$$k_{1} + k_{2} = - p$$
$$k_{1} k_{2} = q$$
$$k_{1} + k_{2} = 4$$
$$k_{1} k_{2} = 3$$
$$k^{2} + k p + q = 0$$
where
$$p = \frac{b}{a}$$
$$p = -4$$
$$q = \frac{c}{a}$$
$$q = 3$$
Vieta Formulas
$$k_{1} + k_{2} = - p$$
$$k_{1} k_{2} = q$$
$$k_{1} + k_{2} = 4$$
$$k_{1} k_{2} = 3$$
Sum and product of roots
[src]
sum
0 + 1 + 3
$$\left(0 + 1\right) + 3$$
=
4
$$4$$
product
1*1*3
$$1 \cdot 1 \cdot 3$$
=
3
$$3$$
3
The graph