Cannot solve quadratic equation through product sum product when the the first product is fraction

Question

I have the folowing equation

$$\frac{16}{3}x^2 - 2x -45 = 0$$ I am trying to solve it for about 40 minutes. I am starting to think I did something wrong. Maybe the equation insn't right, I derived it.

I got as far as this

(16/3)x2 - 2x -45 = 0  (-45 the signs will be different)
               1 . 45
16/3  1        3 . 15
4/3   4        9 . 5
2/3   8

The pairs below the equation are the products. I cannot find the pairs.

You must use the quadratic formula. There is no "nice" factorization. See also WolframAlpha — player3236, Dec 03 '20 at 12:11
I don't like quadratic formula. Not for my classes. The teacher is very good and give us very simple equations. It's almost never necessary to use the complicated quadratic equation. — Diego Alves, Dec 03 '20 at 12:16
@ClaudeLeibovici Can you show me. I new there's a catch in this one! But and I understood, the reason is to eliminate the denominator. — Diego Alves, Dec 03 '20 at 12:18
@DiegoAlves. You got it ! You make you life simpler. Since $3\neq 0$ !! you can do it — Claude Leibovici, Dec 03 '20 at 12:28
@DiegoAlves If you don't like the quadratic formula, you will need to complete the square. To do that, you will need to multiply everything by $3/16$ to make the coefficient of $x^2$ be $1$, and then you have $2bx = -\frac{3}{16}x$ in $(x+b)^2$. — Toby Mak, Dec 03 '20 at 12:29
I got 16x² - 6x - 135 = 0. Is that right? Cannot find the pairs though! — Diego Alves, Dec 03 '20 at 12:39
You can't factorise this equation, so there are no pairs. If you calculate the discriminant, you will find that it is not a perfect square, so the roots are not rational. — Toby Mak, Dec 03 '20 at 12:47

score 2 · Answer 1 · answered Dec 03 '20 at 12:37

$$\frac{16}3x^2 - 2x -45 = 0\\ 16x^2 - 6x -135 = 0\\ \left(4x-\frac34\right)^2 -\frac{3^2}{4^2} -135 = 0\\ \left(4x-\frac34\right)^2 = 135 + \frac{9}{16} = \frac{2169}{16}$$ You can finish the rest yourself. The right hand side is not a nice square, so the solutions are not rational numbers, and it is therefore not surprising that they cannot be found by trying to factorise the original quadratic. Directly applying the quadratic formula is easier than completing the square in a case like this.

Cannot solve quadratic equation through product sum product when the the first product is fraction

1 Answers1