the roots of 2x^2 - 3x = 4 are a and b. find the simplest quadratic equation which has roots 1/a and 1/b

the sum of the roots of the original is 3/2

the product of the roots is -2

so a+b = 3/2
ab = -2

sum of new roots = 1/a + 1/b = (a+b)/(ab)
= (3/2) / -2 = -3/4

product of new roots - (1/a)(1/b) = 1/(ab) = -1/2

so the new equation is
x^2 + (3/4)x - 1/2 = 0 or
4x^2 + 3x - 2 = 0

based on the theorem that for
ax^2 + bx + c = 0

the sum of the roots is -b/a and
the product of the roots is c/a

Why did the quadratic equation go to the circus? Because it wanted to find some roots! Let's get cracking on this problem.

We know that the roots of the equation 2x^2 - 3x = 4 are a and b. That means we can write the equation in factored form as follows:

2x^2 - 3x - 4 = 0
(x - a)(x - b) = 0

To find the equation with roots 1/a and 1/b, we need to flip the roots. So the new roots are a = 1/(1/a) and b = 1/(1/b).

Using the factored form, we can write the new equation as:

(x - 1/a)(x - 1/b) = 0

Now let's simplify it:

(x - 1/a)(x - 1/b) = 0
x^2 - (1/a + 1/b)x + 1/(ab) = 0

Aha! We have our simplest quadratic equation with roots 1/a and 1/b:

x^2 - (1/a + 1/b)x + 1/(ab) = 0

I hope you found this solution both factoring and fact-tastic!

To find the quadratic equation with roots 1/a and 1/b, we need to determine the relationship between the roots of the original equation and the roots of the new quadratic equation.

Let's start by manipulating the equation 2x^2 - 3x = 4:

2x^2 - 3x - 4 = 0

Now we can find the sum and product of the roots using Vieta's formulas:

The sum of the roots (1/a) + (1/b) can be found by utilizing the fact that for a quadratic equation ax^2 + bx + c = 0, the sum of the roots is given by -b/a. In this case, we have:

(1/a) + (1/b) = (-(-3)) / 2 = 3/2

The product of the roots (1/a) * (1/b) can be found using the fact that the product of the roots is c/a. In this case, we have:

(1/a) * (1/b) = (-4) / 2 = -2

Now, we can express the new quadratic equation with roots 1/a and 1/b in the form of:

x^2 - (sum of roots)x + (product of roots) = 0

Substituting the values we found, we have:

x^2 - (3/2)x - 2 = 0

Thus, the simplest quadratic equation with roots 1/a and 1/b is:

x^2 - (3/2)x - 2 = 0.

To find the simplest quadratic equation with roots 1/a and 1/b, we can use the idea that if a polynomial has roots r1 and r2, then their reciprocals 1/r1 and 1/r2 are also roots of the polynomial.

Given that the roots of the quadratic equation 2x^2 - 3x - 4 = 0 are a and b, we can solve for a and b using the quadratic formula:

x = (-b ± √(b^2 - 4ac))/(2a)

In this equation, the coefficient of x^2 is 2, the coefficient of x is -3, and the constant term is -4. Plugging these values into the quadratic formula, we get:

a = (-(-3) ± √((-3)^2 - 4(2)(-4)))/(2(2))
a = (3 ± √(9 + 32))/4
a = (3 ± √41)/4

Similarly, we have:

b = (3 ± √41)/4

Now, let's find the reciprocals of a and b:

1/a = 4/(3 ± √41)
1/b = 4/(3 ± √41)

Since we want to find the simplest quadratic equation, we need to get rid of the denominators in 1/a and 1/b. To do this, we can multiply both the numerator and denominator by the conjugate of the denominator to eliminate the square root:

1/a = (4(3 ∓ √41))/(3^2 - (√41)^2)
= (4(3 ∓ √41))/(9 - 41)
= (4(3 ∓ √41))/(-32)
= -(3 ∓ √41)/8

Similarly, we have:

1/b = -(3 ∓ √41)/8

Now, let's construct the quadratic equation with roots 1/a and 1/b. Since the roots are reciprocals, we have:

(x - 1/a)(x - 1/b) = 0

Expanding this equation, we get:

(x + (3 ∓ √41)/8)(x + (3 ∓ √41)/8) = 0

Simplifying further, we have:

(x + (3 ∓ √41)/8)^2 = 0

Finally, we can distribute and simplify:

x^2 + 2(3 ∓ √41)x/8 + (3 ∓ √41)^2/64 = 0

Thus, the simplest quadratic equation with roots 1/a and 1/b is:

x^2 + 2(3 ∓ √41)x/8 + (3 ∓ √41)^2/64 = 0