I am having a difficult time understanding what I am doing wrong with quadratics. $5x^2-15x-4=0$

Question

I think I am missing a key part of solving quadratic equations. When I solve by grouping, I run into an issue of my values being incorrect, even though I am doing the same thing to both sides of the equation. Could someone explain to me why this doesn't work and what foundation I need to revisit? Here is my work: $$ \begin{align} 5x^2-15x-4&=0\\ 5x^2 -15x &= 4\\ 5x^2–15x-20&=-16\\ 5x^2–20x+5x-20&=-16\\ 5x(x-4)+5(x-4)&=-16\\ (5x+5)(x-4)&=-16\\ \implies x=-4.20, -12 \end{align} $$

Answer 1

Let us follow your approach: till $(5x+5)(x-4) = -16$, all your statements are correct. Now you still need to find which values of $x$ satisfy this, and those are the roots.

However none of the values you mention satisfy - for e.g. taking $x = -12$, this statement reads $(-60+5)(-12-4) = -55\cdot -16 = 880 \neq -16$.

If the RHS of $-16$ factors as $a\times b$ for some $a, b$, and you can find a solution for $x$ which simultaneously satisfies $5x+5=a, x-4=b$, indeed you have a solution. However such factors aren't obvious and in general finding them may be tougher than solving the original quadratic through other methods. Further, you may get only one value of $x$ through this approach, for the other root, you will have to keep checking other possible factorisations.

The only case where you can independently set any one the factors in LHS to equal the RHS to find roots, is when the RHS is $0$, for obvious reasons. This is what you seem to have done, but except when RHS is $0$, this may not solve the original problem. Another e.g. where you may find this working is $(2x-3)(3-x)=1$, due to the coincidence of the factorisation $1^2=1$, even so that gives you only one of the roots.

Hence this approach isn't useful in most cases, and others have suggested valid approaches such as completing the square, or using the quadratic formula, both of which works always.

Answer 2

Your fundamental error is thinking that if $a b = c$ then the two "solutions" are $a = c$ and $b = c$.

Answer 3

Are you trying to complete the square in your problem?

$5x^2-15x-4 = 0$

Then you'd need to do this:

$5x^2-15x = 4$

Then divide by 5:

$x^2-3x=\frac{4}{5}$

Now complete the square:

$x^2-3x + (\frac{-3}{2})^2 = \frac{4}{5} + (\frac{-3}{2})^2$

$(x-\frac{3}{2})^2 = \frac{4}{5}+\frac{9}{4}$

$(x-\frac{3}{2})^2 = \frac{61}{20}$

$(x-\frac{3}{2})^2 = (\sqrt{\frac{61}{20}})^2$

$(x-\frac{3}{2})^2 - (\sqrt{\frac{61}{20}})^2 = 0$

Now you can factor this using $a^2-b^2 = (a-b)(a+b)$

$(x-\frac{3}{2}-\sqrt{\frac{61}{20}})(x-\frac{3}{2} + \sqrt{\frac{61}{20}}) = 0$

Now you can use that ab=0 means a=0 or b=0.

So

$x-\frac{3}{2}-\sqrt{\frac{61}{20}}=0$

or

$x-\frac{3}{2}+\sqrt{\frac{61}{20}}=0$

So solving these two gives

$x = \frac{3}{2} + \frac{1}{2}\sqrt{\frac{61}{5}} \approx3.2464$

or

$x = \frac{3}{2} - \frac{1}{2}\sqrt{\frac{61}{5}} \approx0.2464$

Does this help at all?

I could also have done it like this:

$5x^2-15x = 4$

$5(x^2-3x) = 4$

$5(x^2-3x+(\frac{-3}{2})^2) - 5(\frac{-3}{2})^2 = 4$

$5(x-\frac{3}{2})^2 - \frac{45}{4} = 4$

$5(x-\frac{3}{2})^2 = \frac{61}{4}$

$(x-\frac{3}{2})^2 = \frac{61}{20}$

Then take the square root of both sides. That gives

$x-\frac{3}{2} = \sqrt{\frac{61}{20}}$ or

$x-\frac{3}{2} = -\sqrt{\frac{61}{20}}$

which gives the same two solutions.