Solve x for a quadratic equation (not finding zeroes)

Question

With a linear function $f(x)=5x+2=q$ can be solved for $x$ by rewriting it as $x=(q-2)/5$ While with a quadratic function $f(x)=5x^2+2x+2=q$ how would you solve for both x's on one side? So you clearly can not add the solutions for $5x^2=q$ and $2x+2=q$. Is there an equation to relate y values of $5x^2=$(some relation of q) and $2x+2=$(some relation of q) to q?

Answer 1

Your idea is quite exceptional. It is not generally adopted because the values of x in a quadratic equation can readily be found by methods like factorization, completing square and as well as quadratic formula.

Your suggestion is to break the original into P: 2x + 2 = k and Q : 5x^2 = q – k and hope that the k (if can easily be found) can help in solving the equation with lesser effort. Let see if that is possible or not.

By eliminating x from P and Q, we end up with $5k^2 – 16k + (20 – 4q) = 0$.

Of course, the value(s) of k can be found but we need to solve another quadratic! The worst thing has yet to be come – extraneous value of k must be distinguished and be discarded.

The following uses q = 5 as an example:-

By ‘normal’ methods, we found x = –1 or 3/5.

Using your approach, we have … k = 0 or 16/5.

When k = 0, from P, we get x = –1, but from Q, we get x =–1 or 1(has to be rejected)

When k = 16/5, from P, x = 3/5, but from Q, we get x = 3/5 or –3/5 (has to be rejected)

Note also that the above is just an exceptionally simple example. That means not every quadratic in k has integral/fractional/real roots.

Summing up:- Your approach is a possible way to solve a quadratic but time is wasted in doing many extra work.

Answer 2

So you are asking us to find two convenient functions $f,g$ such that$$5x^2=f(q),\\2x+2=g(q),$$ which are related by $$f(q)+g(q)=q.$$

Obviously, we can eliminate $g(q):=q-f(q)$, and

$$5x^2=f(q),\\2x+2=q-f(q).$$

This system is compatible iff

$$\frac{f(q)}5=\left(\frac{q-f(q)-2}2\right)^2.$$

The solution of this quadratic equation are

$$f(q)=q\pm2\sqrt{\frac q5-\frac9{25}}-\frac85,$$ implying $$g(q)=\mp2\sqrt{\frac q5-\frac9{25}}+\frac85.$$

So yes, it is possible to find these functions, they are well defined, but they don't help making things simpler, on the opposite.

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The way that works is by completing the square $$5x^2+2x+2=5(x+\frac15)^2+2-5\left(\frac15\right)^2=q,$$ then $$\left(x+\frac15\right)^2=\frac q5-\frac9{25}.$$