I am having a hard time with (4a-2x)^2= (5a-3x)(7x-5a). Should I use the Quadratic Formula?
Expanding both sides and the adding/subtracting until the equation = 0 (41a^2+25x^2-66ax = 0). I now have two numbers with an exponent of 2. Using the Quadratic equation example in my book ( ax^2 + bx +c = 0) only has one value with the exponent of two. Am I missing something?
Assuming you are to solve for x, you then write the quadratic in x in standard form:
$\displaystyle 25x^2-66ax+41a^2=0$
To avoid confusion between the a above and the a in the quadratic formula, one might substitute a = k.
$\displaystyle 25x^2-66kx+41k^2=0$
Now identify:
$\displaystyle a=25$
$\displaystyle b=-66k$
$\displaystyle c=41k^2$
Now apply the quadratic formula, and remember to back-substitute for k when you're done.
Edit: You may also observe that:
$\displaystyle 25x^2-66ax+41a^2=0$
is factorable. 25 + 41 = 66...
Yes. Did you notice you could factor the quadratic as:
$\displaystyle (x-a)(25x-41a)=0$
I didn't notice this actually until I saw the discriminant was a perfect square. That's one way to see if a quadratic has rational roots. Then I felt foolish for not noticing that 41 + 25 = 66.