# What will be the $(b^2-a)$?

If $a$, $b$ are real numbers such that $$4a^{2}+b^2=4a-\left(\frac{1}{4b^2}\right)$$ What will be the $b^2-a$ equal to?

I tried to make this equation more basic, but I could not reach the result.

• Is that $\frac{1}{4b^2}$? – almagest Apr 14 '16 at 8:46
• I do not know what is the solution. How did you solve? – ATAKAN İLKGÜN Apr 14 '16 at 8:51
• What is the meaning of (AM/GM) @almagest – ATAKAN İLKGÜN Apr 14 '16 at 8:57
• Arithmetic Mean/Geometric Mean A standard and extremely useful inequality. Given positive $a_1,a_2,\dots,a_n$ we have $\frac{a_1+\dots+a_n}{n}\ge(a_1\dots a_n)^{1/n}$ with equality iff they are all equal. – almagest Apr 14 '16 at 8:57

## 2 Answers

Rearranging we have $(2a-1)^2+(b+\frac{1}{2b})^2=2$. But $|b+\frac{1}{2b}|\ge\sqrt2$ (by AM/GM), so we must have $a=\frac{1}{2},|b|=\frac{1}{\sqrt2}$ and hence $b^2-a=0$.

• Quick and nice! +1 – Macavity Apr 14 '16 at 8:52

With basic tools: $b$ only appears as squares, so $b^2$ can be seen as a positive constant. Remains a quadratic polynomial in $a$: $$4a^2-4a+\left( b^2+\frac{1}{4b^2}\right)=0\,.$$ Its determinant is: $$\Delta = 16\left( 1-\left(b^2+\frac{1}{4b^2}\right)\right) = -16\left(b-\frac{1}{2b}\right)^2\,.$$ Real solutions for $a$ exist only when $|b|=\frac{1}{\sqrt{2}}$, and the double root sums to $1$ (for the coefficient of $a$, negated, divided by the coefficient of $a^2$), hence $a=\frac{1}{2}$, and finally $b^2-a=0$.