Please help me with these three problems!
1. solve for x:
3x + 5x^(1/2) - 28 = 0
2. solve for x:
square root(2x-5) - square root(x-2) = 2
3. Find all values so that polynomial:
ax^2 + 5x + 2 has two distinct real roots.
1. Treat y = sqrt x as the variable ans solve for y. Then factor or use the quadratic equation.
3y^2 + 5y -28 = 0
(3y -7)(y +4) = 0
y = -4 x = +/- 2i are two of the answers. You do the other two.
2. let y^2 = x-2
2x - 5 = 2y^2-1
sqrt (2y^2 -1) = 2 + y
2y^2 -1 = y^2 +4y +4
y^2 -4y -5 = 0
(y-5)(y+1) = 0
y = 5 or -1
Now solve for x
3. Require that b^2 - 4ac > 0
(b = 5 and c = 2).
Solve the inequality for a
Sure! I can help you with these problems. Let's solve them one by one.
1. solve for x: 3x + 5x^(1/2) - 28 = 0
To solve this equation, we can use the quadratic formula. The quadratic formula states that for an equation in the form of ax^2 + bx + c = 0, the solutions for x can be found using the formula:
x = (-b ± √(b^2 - 4ac))/(2a)
In this case, we have 3x + 5x^(1/2) - 28 = 0. To rewrite it in the form ax^2 + bx + c = 0, we can multiply the entire equation by x^(1/2):
3x^(3/2) + 5x - 28x^(1/2) = 0
Now we have an equation in the form ax^2 + bx + c = 0, where a = 3, b = 5, and c = -28. Plugging in these values into the quadratic formula, we get:
x = (-(5) ± √((5)^2 - 4(3)(-28))) / (2(3))
x = (-5 ± √(25 + 336)) / 6
x = (-5 ± √361) / 6
Since √361 = 19, we can simplify the equation further:
x = (-5 ± 19) / 6
So the two solutions for x are:
x₁ = (-5 + 19) / 6 = 14/6 = 7/3
x₂ = (-5 - 19) / 6 = -24/6 = -4
Therefore, the solutions to the equation 3x + 5x^(1/2) - 28 = 0 are x = 7/3 and x = -4.
2. solve for x: √(2x-5) - √(x-2) = 2
To solve this equation, we need to isolate the square roots. Let's start by isolating one of the square roots:
√(2x-5) = √(x-2) + 2
Now square both sides of the equation to eliminate the square roots:
2x-5 = (x-2) + 4√(x-2) + 4
Simplify the equation:
2x-5 = x+2 + 4√(x-2) + 4
Combine like terms:
x - 3 = 4√(x-2)
Isolate the square root term:
4√(x-2) = x - 3
Square both sides of the equation again:
16(x-2) = (x - 3)^2
Multiply out the terms:
16x - 32 = x^2 - 6x + 9
Rewrite the equation in standard form:
x^2 - 22x + 41 = 0
Now you can use the quadratic formula to solve for x:
x = (-(-22) ± √((-22)^2 - 4(1)(41))) / (2(1))
x = (22 ± √(484 - 164)) / 2
x = (22 ± √320) / 2
Simplify further:
x = (22 ± 8√5) / 2
x = 11 ± 4√5
Therefore, the solutions to the equation √(2x-5) - √(x-2) = 2 are x = 11 + 4√5 and x = 11 - 4√5.
3. Find all values so that the polynomial ax^2 + 5x + 2 has two distinct real roots.
For the quadratic polynomial ax^2 + 5x + 2, we can use the discriminant to determine the nature of the roots. The discriminant is calculated as b^2 - 4ac.
In this case, a = a, b = 5, and c = 2. The discriminant is:
D = (5^2) - 4(a)(2)
D = 25 - 8a
To have two distinct real roots, the discriminant must be greater than zero (D > 0):
25 - 8a > 0
To solve for a, we can isolate it:
25 > 8a
a < 25/8
So, all values less than 25/8 (or a < 25/8) will give the quadratic polynomial ax^2 + 5x + 2 two distinct real roots.