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! Please Register or Log in to view the hidden image! 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.
I just remember I learned nothing in my Algebra classes....I know I should be able to do these...yet they seem impossible! Sorry.......Please Register or Log in to view the hidden image!
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 Register or Log in to view the hidden image! Please tell me this is college math Please Register or Log in to view the hidden image!
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. Please Register or Log in to view the hidden image! Might try and get back into algebra. That said, Redrover's explanation went straight over my head. Please Register or Log in to view the hidden image! Please Register or Log in to view the hidden image!
