Certainly! Let's solve the quadratic equation [tex]\( y^2 = 10y + 25 \)[/tex] step-by-step.
First, we'll rearrange the equation to set everything to one side, forming a standard quadratic equation of the form [tex]\( ay^2 + by + c = 0 \)[/tex].
Rearrange the given equation:
[tex]\[ y^2 - 10y - 25 = 0 \][/tex]
Now, let's try to solve this quadratic equation by factoring. For factoring, we need to express the quadratic equation in the form [tex]\( (y - p)(y - q) = 0 \)[/tex] where [tex]\( p \)[/tex] and [tex]\( q \)[/tex] are the roots.
To factor [tex]\( y^2 - 10y - 25 = 0 \)[/tex], we look for two numbers that multiply to give the constant term (-25) and add to give the coefficient of the linear term (-10). However, in this case, it is clear that such numbers do not exist (because the coefficients do not easily factorize). Therefore, factoring by inspection isn't straightforward.
Instead, we apply the quadratic formula:
[tex]\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
In our equation [tex]\( y^2 - 10y - 25 = 0 \)[/tex]:
- [tex]\( a = 1 \)[/tex]
- [tex]\( b = -10 \)[/tex]
- [tex]\( c = -25 \)[/tex]
Now, substitute these values into the quadratic formula:
[tex]\[ y = \frac{-(-10) \pm \sqrt{(-10)^2 - 4 \cdot 1 \cdot (-25)}}{2 \cdot 1} \][/tex]
[tex]\[ y = \frac{10 \pm \sqrt{100 + 100}}{2} \][/tex]
[tex]\[ y = \frac{10 \pm \sqrt{200}}{2} \][/tex]
[tex]\[ y = \frac{10 \pm 10\sqrt{2}}{2} \][/tex]
[tex]\[ y = 5 \pm 5\sqrt{2} \][/tex]
Hence, the solutions to the quadratic equation [tex]\( y^2 = 10y + 25 \)[/tex] are:
[tex]\[ y = 5 - 5\sqrt{2} \][/tex]
[tex]\[ y = 5 + 5\sqrt{2} \][/tex]
So, the solutions to the equation are:
[tex]\[ \boxed{5 - 5\sqrt{2}} \][/tex]
[tex]\[ \boxed{5 + 5\sqrt{2}} \][/tex]
These values represent the points at which the quadratic equation is satisfied.
