How do we get $d$ in terms of $a$ here: $6(d – a)^2 = ad$

Question

I am very sure there is a specific way of solving the equation, $$6(d – a)^2 = ad$$ in order to get $d$ in terms of $a$.

Answer 1

$6(d-a)^2=ad$

$6d^2+6a^2-12ad=ad$

$6d^2-13ad+6a^2=0$

use quadratic formula & solve for d:

$$d=\dfrac{13a\pm\sqrt{169a^2-144a^2}}{12}=\frac{13a\pm 5a}{12}=\dfrac{3a}{2}, \frac{2a}{3}$$ d in terms of a will be, $d=\dfrac{3a}{2}$ or $d=\dfrac{2a}{3}$

Answer 2

Here's my attempt:

First simplify the equation and get all the terms on one side:

\begin{aligned} 6(d-a)^2 = ad \implies 6(d^2-2da+a^2)=ad & \implies 6d^2-12da+6a^2=ad\\ & \implies 6d^2-12da-ad=-6a^2\\ & \implies 6d^2-12ad-ad+6a^2= 0\\ & \implies 6d^2-13ad+6a^2=0 \end{aligned}

Now you can use the quadratic equation and solve for $d$:

$$d = \frac{-\left(-13a\right)\pm \sqrt{\left(-13a\right)^2-4\cdot \:6\left(6a^2\right)}}{2\cdot \:6} \implies d=\frac{13a+5a}{12}=\frac{3a}{2},\:d=\frac{13a-5a}{12}=\frac{2a}{3}$$

Answer 3

Introduce variable $t$, related to $d$ by $d=t+a$. Then you have $$ \begin{align} 6t^2&=at+a^2\\ 6t^2-at&=a^2&&\text{multiply by 24 to make it easy to complete the square}\\ 144t^2-24at&=24a^2\\ 144t^2-24at+a^2&=25a^2\\ (12t-a)^2&=25a^2\\ 12t-a&=\pm5a\\ 12t&=-4a&&\text{ or }12t=6a\\ t&=-\frac13a&&\text{ or }t=\frac12a \end{align} $$

Recalling that $d=t+a$, either $d=\frac23a$ or $d=\frac32a$.

Answer 4

The quadratic formula isn't essential for this particular problem. If we take the relation between the unknowns to be that $ \ d \ $ is some multiple of $ \ a \ $ , then $$ 6·(ka \ - \ a)^2 \ \ = \ \ a·ka \ \ \Rightarrow \ \ 6·(k-1)^2 \ \ = \ \ k \ \ \Rightarrow \ \ 6k^2 \ - \ 13k \ + \ 6 \ \ = \ \ 0 $$ $$ \Rightarrow \ \ k^2 \ - \ \frac{13}{6}k \ + \ 1 \ \ = \ \ 0 \ \ . $$

[The decision to use $ \ d = ka \ $ is a "safe bet" since both sides of the equation are then some multiple of $ \ a^2 \ \ , $ which permits us to "eliminate" $ \ a \ne 0 \ $ from the equation.]

If $ \ k \ $ is a rational number, then we have two solutions $ \ k \ = \ \frac{m}{n} \ $ and $ \ k \ = \ \frac{n}{m} \ \ , $ since the product of the roots for this quadratic equation is $ \ 1 \ \ . $ The sum of the roots is $$ \frac{m}{n} \ + \ \frac{n}{m} \ \ = \ \ \frac{m^2 \ + \ n^2}{mn} \ \ = \ \ \frac{13}{6} \ \ , $$ for which it is fairly clear by inspection that $ \ m \ = \ \pm 3 \ $ and $ \ n \ = \ \pm 2 \ $ (or vice versa). [The choice of rational solutions is justified by the discriminant of the quadratic equation being $ \ \left(-\frac{13}{6} \right)^2 \ - \ 4·1·1 \ = \ \frac{169}{36} \ - \ 4 \ = \ \frac{25}{36} \ \ , $ a "perfect square" of a rational number.]