<li>If 3x^2 2x + 7 = 0, then (x 1/3)^2=</li>
</ol>
<p>A. 20/9
B. 7/9
C. -7/9
D. -8/9
E. -20/9</p>
<p>Anyone know how to work this one out?</p>
<p>you sure you didn't type your choices incorrectly?</p>
<p>i got -20/3 . Im pretty sure thats the answer.</p>
<ol>
<li>If 3x^2 2x + 7 = 0, then (x 1/3)^2=</li>
</ol>
<p>A. 20/9
B. 7/9
C. -7/9
D. -8/9
E. -20/9</p>
<p>Jacked it from a different post in a different forum section- I inquired about the validity of the choices but he/she have not yet responded. Go at it people. </p>
<p>I got -20/3.</p>
<p>Nope the link is right there if you want the legitimate answer</p>
<p>Its E </p>
<p>But I don't have a clue how to work it out</p>
<p>Easy to solve with a graphing calculator.</p>
<p>Wow, I'm out of practice. I don't remember how to solve that simply because the discriminant is -80, meaning the answer is non-real.</p>
<ul>
<li>Oh, nevermind, I just realized it's an x^2 term. Easily done.</li>
</ul>
<p>The answer is x = 1/3 +/- 1.4907i</p>
<p>x - 1/3 = +/- 1.4907i
(x - 1/3)^2 = -(1.4907)^2 = -2.2222 = -20/9</p>
<p>** Here's the quadratic formula solution:</p>
<p>a = 3, b = -2, c = 7</p>
<p>x = (2 +/- sqr(4 - 84)) / 6
x = (2 +/- sqr(-80)) / 6
x = (2 +/- 4rad5i) / 6</p>
<p>x = 1/3 + 2rad5i/3
x = 1/3 - 2rad5i/3</p>
<p>Subtract 1/3:</p>
<p>x - 1/3 = +/-2rad5i/3</p>
<p>Square the x - 1/3 term, and the sign becomes irrelevant (always positive), but i^2 = -1:</p>
<p>(x - 1/3)^2 = 4<em>5</em>-1/9 = -20/9</p>
<p>I think your problem is that when you square 2rad5i/3, you forget to square the denominator.</p>
<p>^ Quix, just use the quadratic formula.</p>
<p>What did you do? I foiled the x-1/3 and eventually came to -7+1/3 , which obviously doesn't add up to -20/9. Explain please</p>
<p>Edit: SAT I requires quadratic formula? thats some bs! is there an alternative?</p>
<p>That's weird; you're working backwards!</p>
<p>Why do you need to expand x-1/3? You just solve for X in the original equation.</p>
<p>That's why you're getting different results.</p>
<p>Quix, could you go over your method? I don't understand how you can make an equality from the first equation and the second, unless you're treating it as a system of equations. I would expect that you'd learn and understand the quadratic equation before doing algebraic systems.</p>
<p>I foiled the X-1/3 and arrive at x^2-2/2X+1/9, then i multiplie it by 3 to get 3x^2-2X+1/3 . </p>
<p>We know that 3x^2-2x+7=0, so 3x^2-2x must be -7.</p>
<p>Now we plug -7 into the foiled equation and get -7+1/3 = -20/3.</p>
<p>Why does that not work?!</p>
<p>The reason it doesn't work is because you're quite randomly multiplying the polynomial x^2 - 2x/3 + 1/9. x^2 - 2x/3 + 1/9 does not equal 3x^2 - 2x + 1/3; you're essentially making up an entirely different set of terms just to fit your convenience. I even take back what I said about having a system of equations: the term (x - 1/3)^2 doesn't equal any given value, so you absolutely have to solve for x before you can solve for (x - 1/3)^2.</p>
<p>Don't plug the objective into the given.</p>
<p>hmm, this is an interesting SAT math problem, usually algebra problems like this can be solved with the use of the quadratic formula. At least from what I've encountered...dunno..</p>
<p>It can be solved with the quadratic formula.</p>
<p>Woops Edit: meant without the use of the quadratic formula.</p>
<p>I think its a SAT2 math prob, SAT I rarely deal with the quadratic formula.</p>
<p>^ I don't think I ever encountered any problem using the quadratic formula. There's almost always an easier solution.</p>
<p>Yeah, thats why its SAT 2</p>
<p>^and because it involves complex numbers, even if implicitly.</p>
<p>Completing the square helps here:
3x^2 – 2x + 7 = 0
3x^2 – 2x = -7
x^2 – 2x/3 = -7/3
x^2 - (2)(x)(1/3) + (1/3)^2 = -7/3 + (1/3)^2
(x - 1/3)^2 = -20/3</p>