Find k in $(1-2k)x^2 - (3k+4)x + 2 = 0$ given facts about the roots.

Question

The exact instruction in my book is:

46. A quadratic equation $(1-2k)x^2 - (3k+4)x + 2 = 0$ is given. Find the value of k for each of the following conditions.

(I got 46a and 46b)

(c) A root of the equation is the negative value of the other root.

(d) A root of the equation is the reciprocal of the other root.

What I tried:

Let $x_1$ and $x_2$ be the roots of the equation.

(c)

$(1-2k)x_1^2 - (3k+4)x_1 + 2 = 0$

$(1-2k)x_2^2 - (3k+4)x_2 + 2 = 0$

$\to (1-2k)x_1^2 - (3k+4)x_1 + 2 = (1-2k)x_2^2 - (3k+4)x_2 + 2$

$\to (1-2k)x_1^2 - (3k+4)x_1 + 2 = (1-2k)x_1^2 + (3k+4)x_1 + 2$

$\to - (3k+4)x_1 = (3k+4)x_1$

$\to x_1 = 0$ or $k = -4/3$

$(x_1 = 0 \to 2 = 0) \to k = -4/3$

Is that right?

(d) The answer given is $k = -1/2$. Wolfram seems to disagree. 1 2

$(1-2k)x_1^2 - (3k+4)x_1 + 2 = 0$

$(1-2k)x_2^2 - (3k+4)x_2 + 2 = 0$

$\to (1-2k)x_1^2 - (3k+4)x_1 + 2 = (1-2k)x_2^2 - (3k+4)x_2 + 2$

$\to (1-2k)x_1^2 - (3k+4)x_1 = (1-2k)x_2^2 - (3k+4)x_2$

$\to (1-2k)x_1^2 - (3k+4)x_1 = (1-2k)(1/x_1)^2 - (3k+4)(1/x_1)$

$\to (1-2k)[x_1^2 - (1/x_1)^2] - (3k+4)[x_1 - (1/x_1)] = 0$

$\to (1-2k)[x_1 - (1/x_1)][x_1 + (1/x_1)] - (3k+4)[x_1 - (1/x_1)] = 0$

$\to (1-2k)[x_1 + (1/x_1)] - (3k+4) = 0$ or $[x_1 - (1/x_1)] = 0$

$\to (1-2k)[x_1 + (1/x_1)] - (3k+4) = 0$, $x_1 = 1$ or $x_1 = -1$

Case 2: $x_1 = 1$

$\to k = -1/5$

Case 3: $x_1 = -1$

$\to k = -7$

Case 1: $(1-2k)[x_1 + (1/x_1)] - (3k+4) = 0$

Case 1a: $x_1 = 1$

$\to k = -2/7$ or $k = 1/2$

Neither k works with $x_1 = 1$

Case 1b: $x_1 = -1$

$\to k = 6$ or $k = 1/2$

Neither k works with $x_1 = -1$

Case 1c: $x_1 \ne -1$, $x_1 \ne 1$

$\{(x_1, k)| x_1 = {\frac{3k + 4 \pm \sqrt{-7k^2 + 40k + 12}}{2-4k}} \}$

Help please? I think I am supposed to somehow prove $x_1 = 1$ or -1 is false and then show $x_1 = \frac{5-i\sqrt{39}}{8}$ or $x_1 = \frac{5+i\sqrt{39}}{8}$

Answer 1

Very good logic! I think you are familiar with the following: If $ax^2+bx+c=0$($a\ne0$)is a quadratic equation and its roots are $\alpha$ and $\beta$,then
$\alpha+\beta=-b/a$
$\alpha \beta=c/a$
Using these,you can easily solve your problem.
Coming to where you probably got stuck:How to refute $x_1=1$ or $-1$?

Dividing given equation by $(1-2k)\ne0$ yields $x^2 - [(3k+4)/(1-2k)]x + 2/(1-2k) = 0$ $\implies$ LHS is a perfect square[both roots are reciprocal so if one root is $1$ other is also $1$]
$x^2 - [(3k+4)/(1-2k)]x + 2/(1-2k) = (x-1)^2=x^2-2x+1$
Comparing coefficients of $x$ on both sides.You get different values of$k$ $\implies$ $x\ne1$
Same process as above can be applied to show:$x\ne-1$

Answer 2

If the roots are distinct, then we can use Vieta's formulas.

(c) One root of the equation is negative of the other. So, sum of the roots is zero. So,$$\frac{3k+4}{1-2k}=0\\ \implies3k+4=0\\ \implies k=-\frac{4}{3}.$$

(d) One root is reciprocal of the other.So, product of the roots is 1. So, $$\frac{2}{1-2k}=1\\ \implies 2=1-2k\\ \implies k=-\frac{1}{2}.$$

The roots are indeed distinct as assuming repeated roots implies contradictions.