find a and b when: x²+8x+21=(x+a)²+b?

Answer 1

x²+8x+21 = (x+a)² + b
x²+8x+16+5 = (x+a)² + b
(x+4)² + 5 = (x+a)² + b
therefore b = 5
(x+a)² = (x+4)²
x + a = x + 4
therefore x =4
~~~

Answer 2

If you complete the square for the trinomial on the left side of the equal sign, you can put it in the vertex form, as expressed on the right side of the equation above.

completing the square (left side expression):

x^2 +8x +21
(x^2 +8x) +21
(x^2 +8x +16) +21 - 16
(x+4)^2 + 5

and setting equal to the right side of your given equation:

(x+4)^2 + 5 = (x+a)^2 + b

And so I would guess that :
a = 4
b = 5

Answer 3

-------- x² + 8x +21 = (x+a)²+ b?
x² + 8x +16 + 5 = (x+a)² + b
(x + 4)² + 5 = (x+a)² + b
a = 4 and b = 5

Answer 4

=(x+4)^2+5 so a=4 and b=5

Answer 5

Note that in general any quadratic polynomial of the form ax^2 + bx + c can be re-written in the form
a(x + b/2a)^2 + (c - b^2/4a), whence we derive the familiar quadratic formula.

So, as already stated, a =4, b = 5

Answer 6

this cannot be solved in this form. The rule is you can only solve for as many variables as you have equations, since you have one equation then you can only solve for one variable. If we assume x to be a known quantity, the best we can do is solve for a or b in terms of the other variable:
x^2+8x+21=x^2+2ax+a^2+b
8x+21=2ax+a^2+b
b= -a^2+(8-2a)x+21

Answer 7

x²+8x+21=(x+a)²+b

x²+8x+21=(x²+2ax+a²)+b
x²+8x+21=x²+2ax+a²+b

8x=2ax
a=4

a²+b=21
(4)²+b=21
b=5


a=4 and b=5