Solve for [tex]$x$[/tex] in the equation [tex]$x^2 - 10x + 25 = 35$[/tex].

A. [tex][tex]$x = 5 \pm 2\sqrt{5}$[/tex][/tex]
B. [tex]$x = 5 \pm \sqrt{35}$[/tex]
C. [tex]$x = 10 \pm 2\sqrt{5}$[/tex]
D. [tex][tex]$x = 10 \pm \sqrt{35}$[/tex][/tex]



Answer :

To solve for [tex]\( x \)[/tex] in the equation [tex]\( x^2 - 10x + 25 = 35 \)[/tex], follow these steps:

1. Rearrange the equation:
[tex]\[ x^2 - 10x + 25 = 35 \][/tex]
Subtract 35 from both sides to set the equation to zero:
[tex]\[ x^2 - 10x + 25 - 35 = 0 \][/tex]
Simplify the equation:
[tex]\[ x^2 - 10x - 10 = 0 \][/tex]

2. Identify the coefficients of the quadratic equation in standard form [tex]\( ax^2 + bx + c = 0 \)[/tex]:
[tex]\[ a = 1, \quad b = -10, \quad c = -10 \][/tex]

3. Use the quadratic formula to solve [tex]\( x \)[/tex]:
The quadratic formula is:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]

4. Calculate the discriminant:
The discriminant [tex]\(\Delta\)[/tex] is given by:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
Plug in the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex]:
[tex]\[ \Delta = (-10)^2 - 4(1)(-10) = 100 + 40 = 140 \][/tex]

5. Compute the two solutions using the quadratic formula:
[tex]\[ x = \frac{-(-10) \pm \sqrt{140}}{2 \cdot 1} = \frac{10 \pm \sqrt{140}}{2} = \frac{10 \pm \sqrt{4 \cdot 35}}{2} = \frac{10 \pm 2\sqrt{35}}{2} \][/tex]
Simplify further:
[tex]\[ x = 5 \pm \sqrt{35} \][/tex]

Therefore, the solutions to the equation [tex]\( x^2 - 10x + 25 = 35 \)[/tex] are:
[tex]\[ x = 10.916079783099615 \quad \text{and} \quad x = -0.9160797830996161 \][/tex]

Thus, [tex]\( x \)[/tex] in the equation [tex]\( x^2 - 10x + 25 = 35 \)[/tex] is:
[tex]\[ x = 5 \pm \sqrt{35} \][/tex]

So, the correct option is:
[tex]\[ x = 5 \pm \sqrt{35} \][/tex]

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