Maths equation?

Question

Don't seem to be able to get the right answer...could someone provide a 'walkthrough'?

x^2+8x+21=(x+a)^2+b

What is the value of a and b?

Answer 1

First, expand the RHS.

(x+a)^2+b
x^2+2ax+a^2+b
x^2+[2a]x+[a^2+b]
Now, it is in the form:
x^2+8x+21
2a = 8
a = 4
a^2+b = 21
(4)^2+b = 21
b = 5

a=4 and b = 5

Hope it helps!

Answer 2

The method for solving this is called 'Completing the Square'.

To find the value for a you have to halve the number in front of the x - this is 8 so the value for a is 4.
So you end up with (x + 4)^2 = x^2 + 8x + 16 when you multiply out the bracket.

You then have to add on 5 to get from 16 to 21 which is in the question. Therefore b = 5

Hope this helps.

Answer 3

First break it up as follows:

(x^2 + 8x) + 21

We know that we can reduce x^2 + 8x to the form (x + 4)^2, however we will be adding 16.

At this point we subtract that 16 from 21 and are left with:

(x + 4)^2 + 5.

Checking the answer gives:

x^2 + 8x + 16 + 5 which equals of course x^2 + 8x + 21

Answer 4

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

Answer 5

x^2+8x+21=x^2+a^2+2ax+b ==> 2ax=8x ==> a=4
a^2 + b=21 ==> 16+b=21 ==> b=5

Answer 6

x^2+8x+21=(x+a)^2+b

expand the term (x + a)^2 + b = x^2 + 2ax + a^2 + b

meaning that the middle term 2ax = 8x where a = 4
and the last term a^2 + b = 21 but a = 4 so b = 5

Answer 7

x^2 + 8x + 21 = (x + a)^2 + b

(x + a)^2 + b = x^2 + 2ax + a^2 + b

Thus, x^2 + 8x + 21 = x^2 + 2ax + a^2 + b
Comparing coefficients of powers of x on both sides of eqn,
2a = 8
a = 4

a^2 + b = 21
4^2 + b = 21
b = 16 - 21 = -5

Answer 8

the given eq. is,

x^2+8x+21=(x+a)^2+b

adding and subtracting 16 in the L.H.S of the eq.

x^2+8x+16 -16 +21 = (x+a)^2 +b

or, (x^2+8x+16) +21-16=(x+a)^2+b

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

therefore,

a=4 and b=5

Answer 9

a and b can be any value if x can be any value.
u need 3 equations for 3 unknowns. Not 1 equation.

Answer 10

a=4
b=5