Sum and product of roots question help please?

Question

Question:

In the quadratic equation (k-2)x^2-5x+2k+3 =0, the roots are reciprocals of each other. Find the value of k.

Thank you for any help! :D

Answer 1

ax² + bx + c = 0

Sum of roots = -b/a

Product of roots = c/a

Roots are reciprocals of each other so c/a = 1

(2k + 3)/(k - 2) = 1

2k + 3 = 1(k - 2)

2k + 3 = k - 2

2k - k = - 2 - 3

k = -5

Answer 2

call the roots a and b so that:

(k-2)(x - a)(x - b) = (k-2)x^2-5x+2k+3

the first bracket of k-2 is included as the coefficient of x^2 id k-2.

the roots are recricals of eachother so:

a = 1/b

(k-2)(x-1/b)(x-b) = (k-2)x^2-5x+2k+3

Now multiply out the left hand side then compare coefficients to create simultaneous equations to find b and k.

Answer 3

A quadratic equation can be written like this:

f(x) = x² - (x₁ + x₂) + x₁x₂ where x₁ and x₂ are the roots.

Notice that your polynomial has a coefficient in front of the x².  Therefore, divide every term by k - 2 to put it in the form given above:

f(x) = x² - [5/(k - 2)]x + [(2k + 3)/(k - 2)]

Therefore, we have that:

x₁x₂ = (2k + 3)/k - 2)

But we are given that the roots are reciprocals of each other which means that x₁x₂ = 1.  Therefore,

1 = (2k + 3)/(k - 2)

k - 2 = 2k + 3

k = -5

Hope this helps!

Answer 4

ax² + bx + c = 0

Sum of roots = -b/a

Product of roots = c/a

Roots are reciprocals of each other so c/a = 1

(2k+3)/(k-2) = 1

2k+3 = k-2

k = -5

Answer 5

(1) First, let's note that the two roots (x1 & x2) are symbolically

as follows:

x1 = [5 + sqrt(y)]/(2k - 4)

x2 = [5 - sqrt(y)]/(2k - 4)

where y = (- 5)^2 - 4(k - 2)(2k + 3) = 25 - 4(k - 2)(2k + 3)

(2) Since x1 = 1/x2, we have:

[5 + sqrt(y)] [5 - sqrt(y)] = 4(k - 2)^2

(3) Expanding the left term, we have:

25 - y = 4(k - 2)^2

(4) Substituting for y [see para (1) above],

25 - y = 25 - [25 - 4(k - 2)(2k + 3)] = 4(k - 2)(2k + 3), so . . .

(5) 4(k - 2)(2k + 3) = 4(k - 2)^2

(6) Dividing both sides by 4(k - 2) gives

2k + 3 = k - 2,

so .... k = - 5

Note: In this last step, dividing by (k - 2) is "legal", since this factor cannot be zero; i.e., k cannot equal 2 because it would violate the quadratic equation constraint.

Answer 6

You mean (2k+3)/k-2 =1,then 2k+3=k-2,then k=-5

======================================...

CHECK: -7 x^2 -5x -7=0,the equation has two imaginary answers.

[-5+(-24)^1/2 j]/-14 & [-5-(-24)^1/2 j]/-14

Answer 7

The product of the roots is 1.

Let the equation be ax^2+bx+c

a=(k-2)

b=-5

c=2k+3

product of the roots = c/a=1

(2k+3)/(k-2) = 1

2k+3 = k-2

k =-5