Which of the following is a solution of [tex]\( x^2 - 7x = -5 \)[/tex]?

A. [tex]\(\frac{7 \pm \sqrt{69}}{2}\)[/tex]

B. [tex]\(\frac{7 \pm \sqrt{29}}{2}\)[/tex]

C. [tex]\(\frac{-7 \pm \sqrt{29}}{2}\)[/tex]

D. [tex]\(\frac{-7 \pm \sqrt{69}}{2}\)[/tex]



Answer :

Given the quadratic equation [tex]\( x^2 - 7x = -5 \)[/tex], let's solve it step-by-step to find the roots and verify which of the provided choices is correct.

Step 1: Rearrange the equation into standard quadratic form

The given equation is:
[tex]\[ x^2 - 7x = -5 \][/tex]

Let's move all terms to one side to set the equation to zero:
[tex]\[ x^2 - 7x + 5 = 0 \][/tex]

Now we have it in the standard quadratic form [tex]\( ax^2 + bx + c = 0 \)[/tex], where:
[tex]\[ a = 1, \quad b = -7, \quad c = 5 \][/tex]

Step 2: Calculate the Discriminant

The discriminant ([tex]\( \Delta \)[/tex]) of a quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex] is given by:
[tex]\[ \Delta = b^2 - 4ac \][/tex]

Substituting the values of [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]:
[tex]\[ \Delta = (-7)^2 - 4(1)(5) = 49 - 20 = 29 \][/tex]

Step 3: Calculate the roots using the quadratic formula

The roots of the quadratic equation can be found using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{\Delta}}{2a} \][/tex]

Substituting the values:
[tex]\[ x = \frac{-(-7) \pm \sqrt{29}}{2(1)} \][/tex]
[tex]\[ x = \frac{7 \pm \sqrt{29}}{2} \][/tex]

Therefore, the solutions to the equation [tex]\( x^2 - 7x + 5 = 0 \)[/tex] are:
[tex]\[ x_1 = \frac{7 + \sqrt{29}}{2} \][/tex]
[tex]\[ x_2 = \frac{7 - \sqrt{29}}{2} \][/tex]

Step 4: Compare with given choices

We can now compare the roots we found with the given options:
1. [tex]\( \frac{7 \pm \sqrt{69}}{2} \)[/tex]
2. [tex]\( \frac{7 \pm \sqrt{29}}{2} \)[/tex]
3. [tex]\( \frac{-7 \pm \sqrt{29}}{2} \)[/tex]
4. [tex]\( \frac{-7 \pm \sqrt{69}}{2} \)[/tex]

The correct choice that matches our calculated roots is:
[tex]\[ \frac{7 \pm \sqrt{29}}{2} \][/tex]

Therefore, the correct answer is:
[tex]\[ \boxed{\frac{7 \pm \sqrt{29}}{2}} \][/tex]