Find the solution(s) to [tex][tex]$x^2-14x+49=0$[/tex][/tex].

A. [tex][tex]$x=7$[/tex][/tex] only
B. [tex][tex]$x=-2$[/tex][/tex] and [tex][tex]$x=7$[/tex][/tex]
C. [tex][tex]$x=7$[/tex][/tex] and [tex][tex]$x=-7$[/tex][/tex]
D. [tex][tex]$x=-1$[/tex][/tex] and [tex][tex]$x=14$[/tex][/tex]



Answer :

To find the solution(s) to the quadratic equation [tex]\( x^2 - 14x + 49 = 0 \)[/tex], let's follow the standard approach using the quadratic formula, which is given by:

[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]

Here, we have the quadratic equation in the form [tex]\( ax^2 + bx + c = 0 \)[/tex], where:
- [tex]\( a = 1 \)[/tex]
- [tex]\( b = -14 \)[/tex]
- [tex]\( c = 49 \)[/tex]

Let’s begin by calculating the discriminant ([tex]\( \Delta \)[/tex]) of the quadratic equation:

[tex]\[ \Delta = b^2 - 4ac \][/tex]

Plug in the values of [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]:

[tex]\[ \Delta = (-14)^2 - 4 \cdot 1 \cdot 49 \][/tex]
[tex]\[ \Delta = 196 - 196 \][/tex]
[tex]\[ \Delta = 0 \][/tex]

The discriminant is 0, which indicates that the quadratic equation has exactly one real solution (a repeated root).

Using the quadratic formula and knowing that [tex]\( \Delta = 0 \)[/tex], we find the solution [tex]\( x \)[/tex] as follows:

[tex]\[ x = \frac{-b \pm \sqrt{0}}{2a} \][/tex]
[tex]\[ x = \frac{-b \pm 0}{2a} \][/tex]
[tex]\[ x = \frac{-(-14)}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{14}{2} \][/tex]
[tex]\[ x = 7 \][/tex]

Therefore, the only solution to the quadratic equation [tex]\( x^2 - 14x + 49 = 0 \)[/tex] is:

[tex]\[ x = 7 \][/tex]

From the given options, the correct answer is:
A. [tex]\( x = 7 \)[/tex] only