1-AND PRODUCTION sum of roots of the equation :X²-X-6=0
2-FOR what values of A the equation has no solution ax ² +3 x-7 = 0
3-For what values of x the equation has 2 solution X²-X+C=0
4-Solve the equation (X-1)²-4=0
5-Solve the equation a)2/2 x^(-3)=2√2
b)3^(x^2-2x)=27
c)5/25 x^(+3)=5√5
help me with the solution of this equations please
1.
Factor the equation to
(x-3)(x+2)=0
So the roots are 3 and -2.
Calculate the sum of roots.
2.
A quadratic equation ax²+bx+c=0 has no real roots when the discriminant is less than zero, that is:
b²-4ac<0
For the present case,
b=3, c=-7
Solve for the inequality in terms of a.
3. and up
Please show some work to indicate where you have difficulties.
a)2/2 x^(-3)=2√2
b)3^(x^2-2x)=27
c)5/25 x^(+3)=5√5
i have difficulties with this three equations
a)
2/(2x^(-3))=2√2
2/(2x^(-3)) = 2√2/1
Cross multiply to get:
2*1 = (2√2)*(2x^(-3))
1=2√2/x³
Cross multiply again:
x³=2√2
x³=(√2)³
Can you take it from here?
b)
3^(x^2-2x)=27
3^(x^2-2x)=3³
therefore
(x^2-2x)= 3
Solve for x by factoring
x²-2x-3=0
(x-3)(x+1)=0
...
c)
similar to a).
Post your answer for checking if you wish.
1) To find the sum of the roots of the equation X² - X - 6 = 0, you can use the Vieta's formulas. Vieta's formulas state that for a quadratic equation of the form ax² + bx + c = 0, the sum of the roots is given by -b/a.
In this case, a = 1 and b = -1. So, the sum of the roots is -(-1)/1 = 1/1 = 1.
Therefore, the sum of the roots of the equation X² - X - 6 = 0 is 1.
2) To determine the values of A for which the equation ax² + 3x - 7 = 0 has no solution, you need to check the discriminant of the quadratic equation. The discriminant, denoted as Δ, is calculated using the formula Δ = b² - 4ac.
In this case, a is the coefficient of the x² term (which is A), b is the coefficient of the x term (which is 3), and c is the constant term (which is -7).
If the discriminant is negative (Δ < 0), then the quadratic equation has no real solutions.
So, for ax² + 3x - 7 = 0 to have no solution, we need the discriminant Δ to be less than 0: Δ = (3)² - 4(A)(-7) < 0.
Simplifying, 9 + 28A < 0.
To find the values of A that satisfy this inequality, solve for A: A < -9/28.
Therefore, for A < -9/28, the equation ax² + 3x - 7 = 0 has no solution.
3) To determine the values of x for which the equation X² - X + C = 0 has 2 solutions, you need to consider the discriminant again. In this case, the discriminant Δ should be greater than 0: Δ = (-1)² - 4(1)(C) > 0.
Simplifying, 1 - 4C > 0.
Solve for C: C < 1/4.
Therefore, for C < 1/4, the equation X² - X + C = 0 has 2 solutions.
4) To solve the equation (X - 1)² - 4 = 0, you need to expand and simplify the equation.
Expanding the square term (X - 1)² gives X² - 2X + 1.
Substituting this back into the equation, we have X² - 2X + 1 - 4 = 0.
Combine like terms to get X² - 2X - 3 = 0.
Now, you can factor or use the quadratic formula to solve for X.
Factoring: (X - 3)(X + 1) = 0
Setting each factor to 0, you get X - 3 = 0 and X + 1 = 0.
Solving for X, you find X = 3 and X = -1.
Therefore, the solutions to the equation (X - 1)² - 4 = 0 are X = 3 and X = -1.
5) Let's solve each part of the equation separately:
a) To solve the equation 2/2x^(-3) = 2√2, first simplify the left-hand side of the equation: 1/x^(-3) = 2√2.
Taking the reciprocal of both sides, you get x^(-3) = 1/(2√2).
Now, raise both sides to the -1/3 power: (x^(-3))^(-1/3) = (1/(2√2))^(-1/3). Simplifying, you get x = ∛(1/(2√2)³).
b) To solve the equation 3^(x^2 - 2x) = 27, first take the logarithm of both sides with respect to base 3.
This gives you x^2 - 2x = log₃(27).
Simplify the right-hand side: log₃(27) = log₃(3³) = 3.
Now, you have the quadratic equation x^2 - 2x - 3 = 0.
Factor or use the quadratic formula to solve for x: (x - 3)(x + 1) = 0.
Setting each factor to 0, you have x - 3 = 0 and x + 1 = 0.
Solving for x, you find x = 3 and x = -1.
c) To solve the equation (5/25)x^3 = 5√5, first simplify the left-hand side: (1/5)x^3 = 5√5.
Multiply both sides by 5 to isolate x^3: x^3 = 25√5.
Now, take the cube root of both sides: ∛(x^3) = ∛(25√5).
Simplifying, you get x = ∛(25√5).
Therefore, the solutions to the equations are:
a) x = ∛(1/(2√2)³)
b) x = 3 and x = -1
c) x = ∛(25√5)