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Should the question say "the equations whose *roots* are the reciprocals of the roots of 2x^2-3x-5 = 0"?

If so, we first want to find the roots of 2x^2-3x-5=0. We factor and find two real roots (2x-5)(x+1).

Set each of these equal to zero to solve for the roots:

2x-5 = 0

2x = 5

x = 5/2

x+1 = 0

x = -1

now we want to find the reciprocals of the roots x = 5/2 and x = -1

Recall that the reciprocal of x = 1/x

Apply this rule to the roots we get

1/x = 1/(5/2) = 2/5 and

1/x = 1/-1 = -1

Now lets find the equation whose roots are equal to 2/5 and -1

We use the equation x^2-Sx+P where S = sum of the roots and P = product of the roots

S = 2/5 - 1 = 2/5 - 5/5 = -3/5

P = -2/5

So, x^2 + 3/5x - 2/5 = 5x^2 + 3x - 2

Check:

5x^2 + 3x - 2 = 0

(x+1)(5x-2) = 0

(x+1) = 0

x = -1

(5x-2) = 0

5x = 2

x = 2/5

QED.

2x^{2}-3x-5 is the equation of a parabola. Where the parabola cuts or touches the x axis are the 'roots'.

we can find these roots if they exist by 3 methods.

1. factoring the equation

2. completing the square

3. using the quadratic formula

by factoring we arrive at (2x-5)(x+1)=0 (multiply the factors together and you will get the original equation)

to get the roots we set each factor = 0

sp 2x-5=0 and x+1=0

which gives us 2x=5 and finally x = 5/2 also x = -1

the reciprocals of the roots are 2/5 and -1

now, if the roots are 2/5 and -1 then (x-2/5) and (x+1) are the factors.

if we multiply them together we get x^{2}+3/5x-2/5)=0 as the needed equation.

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