You can't do that!

In response to Challenge Master's challenge:

Let $a = 8x+13$ , $b=2x-3$ and $c=5$ . From the problem, we have $\begin{cases} \sqrt a + \sqrt b = c &...(1) \\ a + b = c^2 & ...(2) \end{cases}$

We note that $(1)^2 = (2) = c^2$ , therefore,

$\begin{aligned} (\sqrt a + \sqrt b)^2 & = a+b \\ a + 2\sqrt{ab} + b & = a+b \\ \implies ab & = 0 \\ (8x+13)(2x-3) & = 0 \\ \implies x & = \frac 32 = \boxed{1.5} & \small \color{#3D99F6} \text{If } x = - \frac {13}8, \sqrt{2x-3} \text{ is unreal.} \end{aligned}$

Previous solution:

$\begin{aligned} \sqrt{8x+13}+\sqrt{2x-3} &=5 & \small \color{#3D99F6} \text{Let }u^2=2x-3 \\ \sqrt{4u^2+25} + \sqrt{u^2}&=5 \\ \sqrt{4u^2+25} &=5-u \\ 4u^2+25 &= 25-10u +u^2 \\ 3u^2+10u &= 0 \\ u(3u +10) &= 0 \end{aligned}$

$\begin{cases} u=0 & \implies 2x-3=0 & \implies x = \dfrac 32 \\ u=-\dfrac {10}3 & \implies 2x-3= \dfrac {100}9 & \implies x = \dfrac {127}{18} \end{cases}$

Substituting into the original equation, we find that the only acceptable solution is $x =\dfrac 32 =\boxed{1.5}$ .

$\sqrt{8x+13}+\sqrt{2x-3}=5$

or, $(\sqrt{8x+13}+\sqrt{2x-3} )^2=5^2$

or, $8x+13+2x-3+2\sqrt{8x+13}.\sqrt{2x+3}=25$

or, $2\sqrt{8x+13}.\sqrt{2x-3}=15-10x$

or, $(2 \sqrt{8x+13}.\sqrt{2x-3})^{2}=(15-10x)^2$

or, $64x^2-96x+104x-156=225-300x+100x^2$

or, $64x^2-96x+104x+300x-156-225-100x^2=0$

or, $-36x^2+308x-381=0$

or, $36x^2-308x+381=0$ ..........................[multiplying both sides by -1]

$\text{now,calculating that, we get that,}$

$x=\frac{127}{18}, or, x=\frac{3}{2}=1.5$

Since $5$ is the sum of two purely square root numbers. If $0≤ \sqrt{8x+13}≤4$ then $\sqrt{2x-3}$ will be imaginary. So, $3<\sqrt{8x+13}<6$ and $0<\sqrt{2x-3}<5$ as sum is less than $6$ .

If $\sqrt{8x+13}=4$ then $x= \frac{3}{8}$ which is not true for $\sqrt{2x-3}$ (becomes imaginary) and left with $\sqrt{8x+13}=5 \Rightarrow x = \frac{12}{8} = \frac{3}{2}$ which is too true for $\sqrt{2x-3} =0$ .

So the answer is $\dfrac{3}{2} = \boxed{1.5}$

$\sqrt{8x + 13} = 5 - \sqrt{2x - 3}; \\ 8x + 13 = (5 - \sqrt{2x -3})^2; \\ 8x + 13 = 25 - 2x + 3 - 10\sqrt{2x-3}; \\ 10 x - 15 = -10\sqrt{2x-3}; \\ 3 - 2x = 2\sqrt{2x-3}; \\ (3 - 2x)^2 = 2^2(2x - 3); \\ 4x^2 - 12 x + 9 = 8x - 12; \\ 4x^2 - 20 x + 21 = 0; \\ (2x - 3)(2x - 7) = 0; \\ 2x = 3\ \ \text{or}\ \ 2x = 7; \\ x = \frac 3 2 \ \ \text{or}\ \ x = \frac 7 2.$

Check both answers, and we see that only $x = 3/2$ works (after all!)

Why does it work?

In the incorrect, third step, Jason forgot the cross product $2\sqrt{8x+13}\sqrt {2x - 3}$ . However, if $x = 3/2$ then $2x - 3 = 0$ , so that the cross product is zero. By coincidence, leaving out the term had no effect on the sum!

Note that LHS is an increasing function while RHS is constant. Thus there will be atmost one real solution.

Take a=2*x - 3

Now we have, sqrt(4a+25)+sqrt(a)=5=sqrt(25)

It is not hard to guess a=0

Or, x=3/2

I arrived at a quadratic with two solutions 1st X = 7.055555 and X = 1.50 ... But I encounter a strange problem the solution X = 7.055554 = 127/18 is not solution of the proposed problem as it should be ... This no he had never encintrado ... I arrive at the Quad: 36X ^ 2 - 308X + 381 = 0 With solutions: X = 127/18 and X = 3/2 li that is true, but only X = 3/2 is proposed exercise solution .... I can not find any errors ... This should have an explanation ... Can you help me.

Jason got lucky. Define $S= 2x-3$ and then note the following.

$\begin{aligned} 8x + 13 & = 8x -12 + 25 \\ & = 4(2x-3) + 25 \\ & = 4S + 25 \end{aligned}$

So in terms of $S$ , the equation reads $\sqrt{4S + 25} + \sqrt{S} = 5$ .

For the left hand side to be real, $S \geq 0$ . But that implies $\sqrt{4S + 25} \geq \sqrt{0 + 25} = 5$ .

Therefore $\sqrt{4S + 25} + \sqrt{S} \geq 5 + 0$ with equality if and only if $S = 0$ .

Solving $S = 2x-3 = 0$ , we find that $x = \frac{3}{2}$ is the only solution ... just like Jason got in spite of his algebra error.

You could do all the complicated maths, or you could just test the supposedly incorrect answer and find that it proves to be accurate and is therefore the correct answer no matter how you work it out.

(A+B)^2 = A^2 + 2 AB + B^2 = 25 A = (8 x + 13)^0.50 A^2 = (8 x + 13) B = (2 x - 3)^0.50 B^2 = (2 x - 3) 2 AB = 2 * [(8 x + 13)^0.50 * (2 x - 3)^0.50] (A+B)^2 = (8 x + 13) + [2 * (8 x + 13)^0.50 * (2 x - 3)^0.50] + (2 x - 3) 25 = 8 x + 13 + [2 * (16 x^2 + 2 x – 39)^0.50 ] + (2 x - 3) 25 = 10 x + 10 + [2 * (16 x^2 + 2 x – 39)^0.50 ]
25 = 10 x + 10 + [4^0.50 * (16 x^2 + 2 x – 39)^0.50 ]
25 = 10 x + 10 + (64 x^2 + 8 x –156)^0.50
15 – 10 x = (64 x^2 + 8 x –156)^0.50
15 – 10 x = (64 x^2 + 8 x –156)^0.50
100 x^2 -300x + 225 = 64 x^2 + 8 x –156 36x^2 -308x +381 = 0 x^2 - (308/36) x = -381/36 Get complete square by adding square of ½ of the factor of x to both sides x^2 - (308/36) x - (308/72)^2 = -381/36 - (308/72)^2 = -10.583 + 18.299 = 7.716 (x – 308/72)^2 = 7.716 x - 308/72 = +/- (7.716^0.5) x - 308/72 = +/- 2.777 x - 308/72 = + 2.777 + 308/72 = + 2.777 + 4.277 = 7.055 x = - 2.777 + 308/72 = - 2.777 + 4.277 = 1.500 However, only x = 1.5 satisfies the initial expression (8x +13)^0.50 + ()^0.50 = 5 Any value of x <1.5 will give negative value in the second term under the square route on the left hand-side, which is absurd and cannot be accepted Any value of x >1.5 will give a constant on the right hand-side that is >5 and, i.e. violates the initial expression and cannot be accepted

Humm, why does the author of this problem wrote a fallacious reasoning which gives a wrong answer ?

...

Did he dare give us the solution ? Let's try $x = \frac 3 2$ on the equation...

Let's add new variable k = 8x+13,

In than case our equation will look like this:

$\sqrt{k}$ + 1/2* $\sqrt{k-25}$ =5.

K must be equal or more than 25 to prevent us taking square route from negative number.

At the same time k must be less or equal 25, otherwise the sum of 2 square routes will be more than 5.

Only k=25 meets both requirements. That means 8x = 12, x = 1.5