Mathathon 2020 Problem 2 $-$ Easy

Factoring the left-hand side and solving: $\begin{aligned} (4x+1)^2 &= y^2\\ y &= \pm (4x+1)\\ y &= \pm 5 \end{aligned}$

The question should be a bit more specific as to whether the positive or negative root should be accepted. So the answer is technically $\boxed{\pm 5}$ . However, if the question gets edited to dismiss the negative root, we get that the answer is $\boxed{5}$ .

$16x^2 + 8x + 1=y^2$

$(4x + 1)(4x + 1)=y^2$

$(4x + 1)^2=y^2$

$± (4x + 1) = y$

$± 5 = y$

Since $y$ has to be a positive value, $\boxed{y = 5}$

if x = 1:

16 * 1^2 + 8 * 1 + 1 = y^2

25 = y^2

for y >= 0:

y = 5

As a beginning we can rephrase our equation:

$y^2 = 16x^2 + 8x + 1$ is the same as $y = (16x^2 + 8x + 1)^\frac{^1}{2}$

Then we let x equal 1.

$y = (16 \cdot 1^2 + 8 \cdot 1 + 1)^\frac{^1}{2}$

$y = (16 + 8 + 1 )^\frac{1}{2} = \sqrt{25} = 5$

As an alternative we could look at the graph of this function and use our eyes: ;)

Website used for the alternative solution: GeoGebra graphic calculator

$\text{Multiplication by +1 would not change the final output for positive and negative numbers,}$

$\text{so we can just ignore the multiplication altogether}$

$16x^2 + 8x + 1 = y^2$

$16 + 8 + 1 = y^2$

$25 = y^2$

$√25 = y$

$\boxed{5 = y}$

$\begin{aligned} y&=\sqrt{16x^2+8x+1}\\ &=\sqrt{\blue{16x^2+4x}+4x+1}\\ &=\sqrt{4x\blue{(4x+1)}+\blue{4x+1}}\\ &=\sqrt{(4x+1)^2}\\ &=4x+1\\ &=5 \end{aligned}$

$16x^2+8x+1=0\\ x=\cfrac{-8\pm(64-64)}{32}\\ x=0.25\\ (x-0.25)(x-0.25)=0\\ (4x-1)(4x-1)=16\cdot0$

$\text{Break up the equation}$

$\Large{16x^2 + 4x + 4x + 1 = y^2}$

$\text{Factor}$

$\Large{4x(4x + 1) + 1(4x +1) = y^2}$

$\text{Combine}$

$\Large{(4x + 1)^2 = y^2}$

$\text{Remove the powers}$

$\Large{\sqrt{(4x + 1)^2 = y^2}}$

$\text{Substitute}$

$\Large{4(1) + 1 = y}$

$\huge{\boxed{y = 5}}$

$y(>0)=\sqrt {16\times 1^2+8\times 1+1}=\sqrt {25}=\boxed 5$ .

When we factor a polynomial, we can check whether it is in the form $a^2x^2+2ax+1$ . In this question, $a=4$ . This is equal to $\begin{cases} (4x+1)^2=y^2,\\ x=1. \end{cases}$ So the positive value of $y$ is simply $\sqrt{(4x+1)^2}=4x+1=5.$

Just substitute $x=1$ $y^2 = 16(1)^2+8(1)+1$ $\quad=16+8+1\quad\blue{\because 1^2=1}$ $\quad=8(2)+8+1\quad\blue{\because 16=8(2)}$ $=8(2+1)+1=8(3)+1=24+1=25$ $\Rightarrow y^2=25\Rightarrow y={^+_-}\sqrt{25}={^+_-}5$ As $y$ is to be positive $\Rightarrow y=\boxed{5}$

• Any polynomial to be in the form of $(ax + b)^2$ , the polynomial is:

$\begin{aligned} a^2x^2 + 2abx+ b^2 \\16x^2 + 8x + 1 \end{aligned}$

• Here we observe that $a = 4$ AND $b = 1$ .

$16x^2 + 8x + 1 = (4x + 1)^2$

• Given that $x = 1$ , we substitute that and find

$\begin{aligned} y^2 &= (4(1) + 1)^2 \\ y^2 &= 5^2 \\ y^2 - 5^2 &= 0 \\ (y+5)(y-5) &=0 \\ y = \pm 5 \end{aligned}$

• Since it is said to answer as the positive value of $y$ , the answer is

$\boxed{5}$

$y^2=16x^2+8x+1 \implies y^2=16+8+1 (x=1)$ $\implies y^2=25 \implies y =\sqrt{25} = \boxed{\pm 5}$