Mister Exam
Other calculators
-
How to use it?
- Equation:
-
Equation 15x^4-16x^3-30x^2+16x+15=0
- Equation x^3+7x^2-56x+48=0
- Equation 2sin2x–√2·cosx=√2·sinx
- Equation 6t^2-5t-4=0
- Express {x} in terms of y where:
- 17*x+16*y=7
- 17*x-11*y=2
- 15*x-17*y=19
- -12*x+3*y=-9
-
Identical expressions
- 6t^ two -5t- four = zero
- 6t squared minus 5t minus 4 equally 0
- 6t to the power of two minus 5t minus four equally zero
- 6t2-5t-4=0
- 6t²-5t-4=0
- 6t to the power of 2-5t-4=0
- 6t^2-5t-4=O
-
Similar expressions
- 6t^2+5t-4=0
- 6t^2-5t+4=0
6t^2-5t-4=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:
$$t_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$t_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = 6$$
$$b = -5$$
$$c = -4$$
, then
Because D > 0, then the equation has two roots.
or
$$t_{1} = \frac{4}{3}$$
$$t_{2} = - \frac{1}{2}$$
a*t^2 + b*t + c = 0
A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$t_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$t_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = 6$$
$$b = -5$$
$$c = -4$$
, then
D = b^2 - 4 * a * c =
(-5)^2 - 4 * (6) * (-4) = 121
Because D > 0, then the equation has two roots.
t1 = (-b + sqrt(D)) / (2*a)
t2 = (-b - sqrt(D)) / (2*a)
or
$$t_{1} = \frac{4}{3}$$
$$t_{2} = - \frac{1}{2}$$
Vieta's Theorem
rewrite the equation
$$\left(6 t^{2} - 5 t\right) - 4 = 0$$
of
$$a t^{2} + b t + c = 0$$
as reduced quadratic equation
$$t^{2} + \frac{b t}{a} + \frac{c}{a} = 0$$
$$t^{2} - \frac{5 t}{6} - \frac{2}{3} = 0$$
$$p t + q + t^{2} = 0$$
where
$$p = \frac{b}{a}$$
$$p = - \frac{5}{6}$$
$$q = \frac{c}{a}$$
$$q = - \frac{2}{3}$$
Vieta Formulas
$$t_{1} + t_{2} = - p$$
$$t_{1} t_{2} = q$$
$$t_{1} + t_{2} = \frac{5}{6}$$
$$t_{1} t_{2} = - \frac{2}{3}$$
$$\left(6 t^{2} - 5 t\right) - 4 = 0$$
of
$$a t^{2} + b t + c = 0$$
as reduced quadratic equation
$$t^{2} + \frac{b t}{a} + \frac{c}{a} = 0$$
$$t^{2} - \frac{5 t}{6} - \frac{2}{3} = 0$$
$$p t + q + t^{2} = 0$$
where
$$p = \frac{b}{a}$$
$$p = - \frac{5}{6}$$
$$q = \frac{c}{a}$$
$$q = - \frac{2}{3}$$
Vieta Formulas
$$t_{1} + t_{2} = - p$$
$$t_{1} t_{2} = q$$
$$t_{1} + t_{2} = \frac{5}{6}$$
$$t_{1} t_{2} = - \frac{2}{3}$$
Sum and product of roots
[src]
sum
-1/2 + 4/3
$$- \frac{1}{2} + \frac{4}{3}$$
=
5/6
$$\frac{5}{6}$$
product
-4 --- 2*3
$$- \frac{2}{3}$$
=
-2/3
$$- \frac{2}{3}$$
-2/3