$a,b,c,d$ are Distinct reals Find $b+d$

Question

$a,b,c,d$ are Distinct Reals such that: $a,b$ are roots of $x^2-5cx-6d=0$ and $c,d$ are roots of $x^2-5ax-6b=0$. Find Value of $b+d$

My try: we have $a,b,c,d$ as roots of:

$$(x^2-5cx-6d)(x^2-5ax-6b)=0$$ that is $$x^4-5(a+c)x^3+(25ac-6(b+d))x^2+30(bc+ad)x+36bd=0$$

By $Vieta's$ Formulas: $a+b+c+d=5(a+c)$ $\implies$ $$b+d=4(a+c)\tag{1}$$

$$ac=36$$

Also: $$ab=-6d\tag{2}$$ $$cd=-6b\tag{3}$$

From $(1),(2),(3)$ we get:

$$b+d=-24\left(\frac{b}{d}+\frac{d}{b}\right)$$

That is:

$$bd=\frac{24(b+d)^2}{48-(b+d)}\tag{A}$$

Also by $Vieta's$ Sum of bi quadratic roots taken two at a time we get: $$ac+a(b+d)+c(b+d)+bd=25ac-6(b+d)$$

$\implies$ $$36+(a+c)(b+d)+bd=900-6(b+d)$$

From $(1)$ we get:

$$\frac{(b+d)^2}{4}+6(b+d)=864-bd \tag{B}$$

Letting $b+d=y$ we get from $(A)$ :

$$\frac{y^2}{4}+6y=864-\frac{24y^2}{48-y}$$

Solving this we get:

$y=144$ or $y=24$ or $y=-48$

Any better approach?

Answer 1

Perhaps with Vieta formula's for a quadratic equation would go faster:

$$ a+b=5c$$ $$ab=-6d$$ $$c+d=5a$$ $$cd=-6b$$

So we have $$ {a+b\over 5}-{ab\over 6}=4a\implies 6b-5ab =114a$$ and $${a+b\over 5}\cdot{ab\over 6} =6b\implies a(a+b)=180\;\;\vee \;\; b=0$$

• If $b=0$ we get $a=0$ so impossible.
• If $a\ne 6/5$ then $b ={114a\over 6-5a}$ and thus we have to solve: $$a^2+{114a^2\over 6-5a} = 180$$ which is the same as $$a^3-24a^2-180a+216 =0$$

Answer 2

As we need to eliminate $a,c$

use $ab=-6d,cd=-6b$

$5c=a+b\implies-30b/d=-6d/b+b\iff30b^2+b^2d=6d^2$

Similarly $5a=c+d\implies-30d/b=-6b/d+d\iff6b^2=30d^2+bd^2$

$6(d^2-b^2)=-30(d^2-b^2)-bd(d-b)$

$36(d+b)=bd\ \ (1)$

Also $b^2(30+d)=6d^2,d^2(30+b)=6b^2$

$36=(30+b)(30+d)=900+30(b+d)+bd\ \ (2)$

Replace the value of $bd$ from $(1)$ in $(2)$

Answer 3

As we need to eliminate $a,c$

use $a+b=5c, 5a=c+d\implies24a=b+5d,24c=5b+d$

$$-6d=ab=\dfrac{b(b+5d)}{24}\iff b^2+5bd=-144d$$

$$-6b=cd=\dfrac{d(5b+d)}{24}\iff d^2+5bd=-144b$$

Subtract to find $$(b-d)(b+d)=144(b-d)$$