# (Pretty Basic) Algebra Problems.

Registered Senior Member
First Problem:

3<i>x</i><sup>2</sup> + 12<i>x</i> +5 = <i>p</i>(<i>x</i> + <i>q</i>)<sup>2</Sup> + <i>r</i>

For all values of <i>x</i>

Find <i>p</i>, <i>q</i> and <i>r</i>. And find the "minimum value" of 3<i>x</i><sup>2</sup> + 12<i>x</i> +5.

Second problem:

Find <i>a</i>, <i>b</i> and <i>c</i> such that

<i>a</i>(<i>x</i><sup>2</sup> + 4) + (<i>bx</i> + <i>c</i>) is identical to 7<i>x</i><sup>2</sup> - <i>x</i> + 14

------

Can anyone help with these?

The first one is a peice of cake.

p(x + q)² + r = p(x² + 2xq + q²) + r
= px² + 2pqx + pq² + r

You can then say that:
3x² = px²
therefore:
3 = p

then:

2pqx = 12x
q= 12x / 2px = 12 / 2*3 [remember, p = 3]
q = 2

finally:

pq² + r = 5
3 * 2² + r = 5
12 + r = 5
r = 5 - 12 = -7

Minumum value:
That one is thougther. Luckely, we now have the canonical form (ei p(x + q)² + r = 3(x + 2)² - 7).
So the lowest point of the quadriatic equation is whenever 3(x + 2)² = 0, therefore when x = -2

So the minimum value is 3(-2 + 2)² - 7 = -7

The second one is a bit more of a pain.

a(x² + 4) + (bx + c) = ax² + 4a + bx + c = 7x² - x + 14

So:

ax² = 7x²
bx = -1 x
and
4a + c = 14 = 4*7 + c
c = 14 - 28 = -14

Thanks for the help!

But, you know what? I made an error with the second problem. it <i>should</i> read:

<i>a</i>(<i>x</i><sup>2</sup> + 4) +<b>(<i>x</i>-2)</b>(<i>bx</i> + <i>c</I>)

Sorry for the mistake. But your first effort was not in vain. It's given me a better Idea of how to solve these.

Last edited:
The answer is a=5, b=2, c=3

I just remember I learned nothing in my Algebra classes....I know I should be able to do these...yet they seem impossible! Sorry.......

I see.

(a + b)x<sup>2</sup> + (c - 2b)x + 4a + 2c = 7x<sup>2</sup> - x + 14

Thus

(a + b) = 7

(c - 2b) = -1

(4a - 2c) = 14

But.

What step did you then take to deduce that a = 5, b= 2 and c = 3 ? Other than trial and error? I just don't see it.

a + b = 7 ...[1]

c - 2b = -1 ...[2]

4a - 2c = 14 ...[3]

You need to solve these simultaneously.

Divide [3] by 2:

2a - c = 7 ...[3b]

Rearrange [1] and [2]:

a = 7 - b ...[1b]
c = 2b - 1 ...[2b]

Plug [1b] and [2b] into [3b]:

2(7-b) - (2b -1) = 7

Solve:

14 - 2b - 2b + 1 = 7

4b = 8

b = 2

From [1b] and [2b]:

a = 7 - 2 = 5
c = 2.2 - 1 = 3

So a=5, b=2, c=3

AD1, is that college algebra ? I've never seen anything close to
this in my high school advanced math class... aside from equations
and inequations.

Please tell me this is college math

Thanks for the help James R.

Soliddus, this is high school level over in the UK.

As Cartman says: "Ahh, SON OF A BITCH!"

I used a matrix to find my answer but James R's solution is probably the one your teacher is looking for.

This is frankly 10th grade math. If this is thought, you should see my calculus class.

I hate algebra, but there's something very satisfying when you finish, and get the rght answer. Might try and get back into algebra.

That said, Redrover's explanation went straight over my head.