Answer :
To solve the quadratic equation [tex]\(-x^2 + 8x = -15\)[/tex], we first need to rewrite it in standard form.
Starting from:
[tex]\[ -x^2 + 8x = -15, \][/tex]
we move the constant term to the left side to get:
[tex]\[ -x^2 + 8x + 15 = 0. \][/tex]
We multiply through by -1 to get the canonical form of a quadratic equation:
[tex]\[ x^2 - 8x - 15 = 0. \][/tex]
Next, we use the quadratic formula to find the roots of the equation of the form [tex]\( Ax^2 + Bx + C = 0 \)[/tex]:
[tex]\[ x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}. \][/tex]
For the given equation [tex]\( x^2 - 8x - 15 = 0 \)[/tex]:
- [tex]\( A = 1 \)[/tex]
- [tex]\( B = -8 \)[/tex]
- [tex]\( C = -15 \)[/tex]
The discriminant [tex]\( D \)[/tex] is calculated as follows:
[tex]\[ D = B^2 - 4AC = (-8)^2 - 4(1)(-15) = 64 + 60 = 124. \][/tex]
Substituting [tex]\( B \)[/tex], [tex]\( D = 124 \)[/tex], and [tex]\( A = 1 \)[/tex] into the quadratic formula gives us:
[tex]\[ x = \frac{-(-8) \pm \sqrt{124}}{2(1)}. \][/tex]
[tex]\[ x = \frac{8 \pm \sqrt{124}}{2}. \][/tex]
Simplify further:
[tex]\[ x = \frac{8 \pm 2\sqrt{31}}{2}. \][/tex]
[tex]\[ x = 4 \pm \sqrt{31}. \][/tex]
So, the values of [tex]\( x \)[/tex] that satisfy the equation [tex]\(-x^2 + 8x = -15\)[/tex] are:
[tex]\[ x = 4 \pm \sqrt{31}. \][/tex]
Thus, the correct answer is:
C. [tex]\( 4 \pm \sqrt{31} \)[/tex].
Starting from:
[tex]\[ -x^2 + 8x = -15, \][/tex]
we move the constant term to the left side to get:
[tex]\[ -x^2 + 8x + 15 = 0. \][/tex]
We multiply through by -1 to get the canonical form of a quadratic equation:
[tex]\[ x^2 - 8x - 15 = 0. \][/tex]
Next, we use the quadratic formula to find the roots of the equation of the form [tex]\( Ax^2 + Bx + C = 0 \)[/tex]:
[tex]\[ x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}. \][/tex]
For the given equation [tex]\( x^2 - 8x - 15 = 0 \)[/tex]:
- [tex]\( A = 1 \)[/tex]
- [tex]\( B = -8 \)[/tex]
- [tex]\( C = -15 \)[/tex]
The discriminant [tex]\( D \)[/tex] is calculated as follows:
[tex]\[ D = B^2 - 4AC = (-8)^2 - 4(1)(-15) = 64 + 60 = 124. \][/tex]
Substituting [tex]\( B \)[/tex], [tex]\( D = 124 \)[/tex], and [tex]\( A = 1 \)[/tex] into the quadratic formula gives us:
[tex]\[ x = \frac{-(-8) \pm \sqrt{124}}{2(1)}. \][/tex]
[tex]\[ x = \frac{8 \pm \sqrt{124}}{2}. \][/tex]
Simplify further:
[tex]\[ x = \frac{8 \pm 2\sqrt{31}}{2}. \][/tex]
[tex]\[ x = 4 \pm \sqrt{31}. \][/tex]
So, the values of [tex]\( x \)[/tex] that satisfy the equation [tex]\(-x^2 + 8x = -15\)[/tex] are:
[tex]\[ x = 4 \pm \sqrt{31}. \][/tex]
Thus, the correct answer is:
C. [tex]\( 4 \pm \sqrt{31} \)[/tex].