The problem provides an incomplete table of values for the quadratic function $y = x^2 - 4x + 1$. The questions are: (i) Find the value of $y$ when $x = 2$. (ii) Draw the graph of the quadratic function using the given table and a suitable scale. (iii) Write the range of $x$ for which $y < 0$. (iv) Express the function in the form $y = (x-a)^2 + b$, where $a$ and $b$ are constants. (v) By considering the positive root of the equation $1 - 4x + x^2 = 0$, find the value of $\sqrt{3}$.

1. Problem Description

The problem provides an incomplete table of values for the quadratic function $y = x^2 - 4x + 1$. The questions are:

(i) Find the value of $y$ when $x = 2$.

(ii) Draw the graph of the quadratic function using the given table and a suitable scale.

(iii) Write the range of $x$ for which $y < 0$.

(iv) Express the function in the form $y = (x-a)^2 + b$, where $a$ and $b$ are constants.

(v) By considering the positive root of the equation $1 - 4x + x^2 = 0$, find the value of $\sqrt{3}$.

2. Solution Steps

(i) Find the value of $y$ when $x = 2$.

Substitute $x = 2$ into the equation $y = x^2 - 4x + 1$:

$y = (2)^2 - 4(2) + 1 = 4 - 8 + 1 = -3$

So, the missing value in the table when $x=2$ is $-3$.

(ii) Draw the graph of the quadratic function using the given table and a suitable scale.

The completed table is:

x | -1 | 0 | 1 | 2 | 3 | 4 | 5

---|---|---|---|---|---|---|---

y | 6 | 1 | -2 | -3 | -2 | 1 | 6

Plot the points on a graph paper and draw a smooth curve through the points. We can't show the graph here, but imagine plotting the points and connecting them.

(iii) Write the range of $x$ for which $y < 0$.

From the graph (or the table), $y < 0$ when $x$ is between approximately 0.27 and 3.

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3. Therefore, the range is $0.27 < x < 3.73$.

(iv) Express the function in the form $y = (x-a)^2 + b$, where $a$ and $b$ are constants.

Complete the square for the given equation $y = x^2 - 4x + 1$.

$y = x^2 - 4x + 4 - 4 + 1$

$y = (x - 2)^2 - 3$

So, $a = 2$ and $b = -3$.

(v) By considering the positive root of the equation $1 - 4x + x^2 = 0$, find the value of $\sqrt{3}$.

We are given the equation $x^2 - 4x + 1 = 0$, or $1-4x+x^2 = 0$.

From part (iv), we have $y = (x - 2)^2 - 3$. The roots of the equation $(x - 2)^2 - 3 = 0$ are the values of x where y =

0. So, $(x - 2)^2 = 3$.

Taking the square root of both sides:

$x - 2 = \pm\sqrt{3}$

$x = 2 \pm \sqrt{3}$

The positive root is $x = 2 + \sqrt{3}$.

We are also given $1-4x+x^2 = 0$. Thus, $x^2-4x+1=0$. Therefore the positive root from the equation can be calculated using quadratic formula as follows

$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} = \frac{4 \pm \sqrt{16-4}}{2} = \frac{4 \pm \sqrt{12}}{2} = \frac{4 \pm 2\sqrt{3}}{2} = 2 \pm \sqrt{3}$.

The positive root is $2 + \sqrt{3}$.

From part (v), $1 - 4x + x^2 = 0$, so $x = 2 + \sqrt{3}$. We are to derive $\sqrt{3}$.

We have $x=2 + \sqrt{3}$, so $\sqrt{3} = x - 2$.

Now, given the equation is $1-4x + x^2 = 0$. Let us compare this with the original equation $y = x^2-4x+1 = 0$. Thus, $y = 0$. Now look at expression in the form $(x-2)^2-3 = 0$. $(x-2)^2 = 3$, implies $x-2 = \sqrt{3}$. So $x = \sqrt{3} + 2$. But $y=0$. And the values of $x$ when $y=0$, by using the graph, $x=0.27$ or $x= 3.73$. Thus $x= 2+\sqrt{3}$. That is $x = 2 + 1.73$. That gives $x=3.73$. And we already know $x = 2 + \sqrt{3}$, so $\sqrt{3} = x-2$. From graph $x=3.73$ so $\sqrt{3} = 3.73-2=1.73$.

3. Final Answer

(i) $y = -3$

(ii) The graph is a parabola passing through (-1,6), (0,1), (1,-2), (2,-3), (3,-2), (4,1), (5,6).

(iii) $0.27 < x < 3.73$ (approximately)

(iv) $y = (x - 2)^2 - 3$

(v) $\sqrt{3} = 1.73$ (approximately)