Solve the following equation: $\sqrt{4x^2 - 15x + 20} = 4x - 10 + 7\sqrt{x - 1}$.

Question

Solve the following equation. $$\large \sqrt{4x^2 - 15x + 20} = 4x - 10 + 7\sqrt{x - 1}$$

I try to let $2x - 6 = a$ and $\sqrt{x - 1} = b$. Then the equation becomes:

$$\sqrt{a^2 + 9b^2 - 7} = 2a + 7b + 2$$

with which I think about how stupid I am about what I am working.

Notation: I am trying to do a solution involving inequalities. So I will be grateful if anyone can do that.

Answer 1

Letting $y = \sqrt{x-1}$, your equation is equivalent to $$\sqrt{4(y^2+1)^2-15(y^2+1)+20} = 4(y^2+1)-10 + 7y$$

i.e. $$\sqrt{4y^4 - 7y^2 + 9} = 4y^2 + 7y - 6$$

so squaring it, you get $$4y^4 - 7y^2 + 9 = 16y^4 + 56y^3 + y^2 - 84y + 36$$

i.e. $$12y^4 + 56y^3 + 8y^2 - 84y + 27 = 0$$

i.e. $$(2y^2+2y-3)(6x^2+22x-9)=0$$

You get that $$y= \frac{1}{6} \left(- 11 - 5 \sqrt{7} \right) \quad \text{or} \quad y= \frac{1}{2} \left(- 1 - \sqrt{7} \right) \quad$$ $$\text{or} \quad y= \frac{1}{2} \left(- 1 + \sqrt{7} \right) \quad \text{or} \quad y= \frac{1}{6} \left(- 11 + 5 \sqrt{7} \right) \quad $$

The first two ones are impossible because $y$ has to be positive, and the last one is impossible (because $4y^2+7y-6$ has to be positive) so you get $$\quad y= \frac{1}{2} \left(- 1 + \sqrt{7} \right)$$

i.e. $$x= \left(\frac{1}{2} \left(- 1 + \sqrt{7} \right)\right)^2+1 \quad \text{i.e. } \quad x= 3 - \frac{1}{2} \sqrt{7}$$

Answer 2

Let $t= \sqrt{x-1}\geq 0$ then we get $$\sqrt{4t^4 - 7t^2 + 9} = 4t^2+7t-6$$

and now square it, we get $$12t^4+56t^3+8t^2-84t+27=0$$

Now use divide this by $t^2$ (clearly $t \ne 0$) and we get $$3(4t^2+{9\over t^2}) -28(2t-{3\over t})+8=0$$

Let $y=2t-{3\over t}$ and we get $$3(y^2+12)-28y+8=0$$ so $$3y^2-28y+44=0$$

now solve this and then finish...

Answer 3

Hint: Squaring one times we get $$-12x^2+16x-31=14(4x-10)\sqrt{x-7}$$ Squaring again and simplifying and factorizing we get $$(36x^2-664x+709)(4x^2-24x+29)=0$$

Answer 4

The domain gives $x\geq1$.

Thus, for all root of the equation we obtain: $$4x-10+7\sqrt{x-1}=\sqrt{4x^2-15x+20}=\sqrt{x^2+3x^2-15x+20}>x,$$ which gives $$3x+7\sqrt{x-1}>10.$$

But for $x=1.5$ we see that $$3x+7\sqrt{x-1}=4.5+\frac{7}{\sqrt2}<10$$ and since $3x+7\sqrt{x-1}$ increases, we obtain $x>1.5.$

In another hand, for $x>1.5$ we obtain: $$4x-10+7\sqrt{x-1}=\sqrt{4x^2-15x+20}<2x,$$ which gives $$2x+7\sqrt{x-1}<10.$$

Now, for $x=2$ we see that $$2x+7\sqrt{x-1}=4+7>10$$ and since $2x+7\sqrt{x-1}$ increases, we obtain $x<2$.

Now, rewrite our equation in the following form: $$\sqrt{4x^2-15x+20}-3\sqrt{x-1}=2(2x-5+2\sqrt{x-1})$$ or $$\frac{4x^2-24x+29}{\sqrt{4x^2-15x+20}+3\sqrt{x-1}}=\frac{2(4x^2-24x+29)}{2x-5-2\sqrt{x-1}},$$ which gives $$4x^2-24x+29=0$$ and since $1.5<x<2,$ we obtain $$x=3-\frac{\sqrt7}{2}$$ or $$\frac{1}{\sqrt{4x^2-15x+20}+3\sqrt{x-1}}=\frac{2}{2x-5-2\sqrt{x-1}},$$ which gives $$2\sqrt{4x^2-15x+20}+8\sqrt{x-1}=2x-5,$$ which is impossible because $$2\sqrt{4x^2-15x+20}+8\sqrt{x-1}>2x>2x-5.$$ Id est, we got the following answer: $$\left\{3-\frac{\sqrt7}{2}\right\}.$$