Answer :
[tex]b^2-4b+4=0[/tex]
Before using the quadratic, make sure the right side is equal to 0. Check.
The quadratic formula gives the solution for an equation [tex]ax^2+bx+c=0[/tex],
which is [tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex].
In this case, our "a" value can be considered 1, b = -4, and c = 4.
Let's plug these values into the quadratic formula.
[tex]=\frac{-(-4)\pm\sqrt{(-4)^2-4(1)(4)}}{2(1)}=\frac{4\pm\sqrt{16-16}}2=\frac{4\pm\sqrt0}2=\frac{4}2=\boxed{2}[/tex]
Another easy way to solve would be by factoring.
We would find two numbers that add to b and mutliply to equal ac.
Split our bx term into these two values
Factor and simplify.
(These numbers would be -2 and -2. Then we'd get
b² - 2b - 2b + 4. Factor the first two terms...
b(b-2) - 2b + 4. Factor the last two so that you get the same thing in the ()'s
b(b-2) -2(b-2) Combine like terms.
(b-2)(b-2) or (b-2)² is our factored form.
This is going to be equal to zero, so (b-2)² = 0.
We then look at our factors and note that any number which can cause it to equal zero is correct. Both factors are (b-2). In that case, b=2.
There's also a lot of shortcuts you could've done there such as taking the two numbers you got (say, m and n) and putting them directly into (x+n)(x+m) whenever your a value is 1.)
Before using the quadratic, make sure the right side is equal to 0. Check.
The quadratic formula gives the solution for an equation [tex]ax^2+bx+c=0[/tex],
which is [tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex].
In this case, our "a" value can be considered 1, b = -4, and c = 4.
Let's plug these values into the quadratic formula.
[tex]=\frac{-(-4)\pm\sqrt{(-4)^2-4(1)(4)}}{2(1)}=\frac{4\pm\sqrt{16-16}}2=\frac{4\pm\sqrt0}2=\frac{4}2=\boxed{2}[/tex]
Another easy way to solve would be by factoring.
We would find two numbers that add to b and mutliply to equal ac.
Split our bx term into these two values
Factor and simplify.
(These numbers would be -2 and -2. Then we'd get
b² - 2b - 2b + 4. Factor the first two terms...
b(b-2) - 2b + 4. Factor the last two so that you get the same thing in the ()'s
b(b-2) -2(b-2) Combine like terms.
(b-2)(b-2) or (b-2)² is our factored form.
This is going to be equal to zero, so (b-2)² = 0.
We then look at our factors and note that any number which can cause it to equal zero is correct. Both factors are (b-2). In that case, b=2.
There's also a lot of shortcuts you could've done there such as taking the two numbers you got (say, m and n) and putting them directly into (x+n)(x+m) whenever your a value is 1.)
