Question

X^2-4x+1 draw the following find the zeros

Answer 1

Answer:

$\large\boxed{\sf{(2+\sqrt{3})\;\;and\;\;(2-\sqrt{3})}}$

Step-by-step explanation:

Given a quadratic polynomial such that,

${x}^{2} - 4x + 1$

Now, we know that, general form a quadratic polynomial is given by ,

$a {x}^{2} + bx + c$

Comparing both the forms, we get,

• a = 1
• b = -4
• c = 1

Drawing the graph :-

Here, we have, a > 0

We know that, for a > 0, graph is upward parabola.

Also, we know that, descriminant, D is given by,

• $\large \bold{{b}^{2} - 4ac}$

Therefore, we will get,

$=>D ={(-4)}^{2}-4(1)(1)\\\\=> D = 16 - 4\\\\=> D = 12$

Clearly, D > 0.

Therefore, the roots will be real and distinct.

Therefore, the graph will cut the X axis at two distinct points.

Also, the lower most point of graph is given by,

• $\large\bold{( \dfrac{ - b}{2a} , \dfrac{-D}{4a} )}$

Therefore, the lower most point is (2,-3)

$\red{Note:-}$ Refer to attachment for graph.

Finding the roots :-

To find the roots, equate the function to 0.

Therefore, we will get,

$= > {x}^{2} - 4x + 1 = 0 \\ \\ = > x = \frac{ - ( - 4) \pm \sqrt{ 12 } }{2} \\ \\ = > x = \frac{4 \pm2 \sqrt{3} }{2} \\ \\ = > x = 2 \pm \sqrt{3}$

Hence, the roots are (2 + √3) and (2 - √3).