1. ## Finding the roots of a Quadratic equation.

I am having a hard time with (4a-2x)^2= (5a-3x)(7x-5a). Should I use the Quadratic Formula?

2. ## Re: Finding the roots of a Quadratic equation.

You should expand both sides of the equation, then collect like terms with one side being zero first.

3. ## Re: Finding the roots of a Quadratic equation.

Originally Posted by Berean
I am having a hard time with (4a-2x)^2= (5a-3x)(7x-5a). Should I use the Quadratic Formula?
There is no common factor so, yes, go ahead and multiply out each side, put the quadratic in "standard form" and either factor (if you can), complete the square, or use the quadratic formula.

4. ## Re: Finding the roots of a Quadratic equation.

Originally Posted by MarkFL2
You should expand both sides of the equation, then collect like terms with one side being zero first.
Thankyou

Thankyou

6. ## Re: Finding the roots of a Quadratic equation.

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?

7. ## Re: Finding the roots of a Quadratic equation.

Assuming you are to solve for x, you then write the quadratic in x in standard form:

$25x^2-66ax+41a^2=0$

To avoid confusion between the a above and the a in the quadratic formula, one might substitute a = k.

$25x^2-66kx+41k^2=0$

Now identify:

$a=25$

$b=-66k$

$c=41k^2$

Now apply the quadratic formula, and remember to back-substitute for k when you're done.

Edit: You may also observe that:

$25x^2-66ax+41a^2=0$

is factorable. 25 + 41 = 66...

8. ## Re: Finding the roots of a Quadratic equation.

I have come up with an answer of .....a,(41/25)a. So the roots of (4a-2x)^2 = (5a-3x)(7x-5a) is a,(41/25)a?

9. ## Re: Finding the roots of a Quadratic equation.

Yes. Did you notice you could factor the quadratic as:

$(x-a)(25x-41a)=0$

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.

10. ## Re: Finding the roots of a Quadratic equation.

Originally Posted by MarkFL2
Yes. Did you notice you could factor the quadratic as:

$(x-a)(25x-41a)=0$

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.
You will feel even more foolish when you realise 66 is not a square number although the discriminant is a square

11. ## Re: Finding the roots of a Quadratic equation.

Originally Posted by e^(i*pi)
You will feel even more foolish when you realise 66 is not a square number although the discriminant is a square

Won't you feel foolish when you realise you missed the point of 25 + 41 = 66

??