Real ordered pair $(a,b)$ in equation

Question

If $$a^2+5b^2+2b=6a+2ab-10$$

then all real ordered pair of $(a,b)$ is?

Try: I am trying to convert it into $$(..)^2+(..)^2=0$$ or $$(..)^2+(..)^2+(..)^2=0.$$

So $$(a-b)^2+(2a-0.75)^2+10-(0.75)^2-6a=0.$$

Could some help me to solve it? Thanks.

Answer 1

$a^2+5b^2+2b-6a-2ab+10=0$

Let's suppose we had a $(a+kb+j)^2$ term and that were the only term with $a $. We'd have $(a+kb+j)^2=a^2+k^2b^2+j^2+2kab+2aj+2kjb $. So $k=-1;j=-3$.

So

$a^2+5b^2+2b-6a-2ab+10=0$

$a^2+b^2+9-2ab-6a+6b+4b^2-4b+1=0$

$(a-b-3)^2+(2b-1)^2=0$

So for real solutions

$2b-1=0$ and $b=\frac 12$

And $a-b-3=0$ so $a=3\frac 12$.

Answer 2

$$ ( a-b-3 )^2 + (2b-1)^2 = a^2 -2ab+5b^2 -6a +2b + 10 $$

In what follows, $H$ is half the Hessian matrix of $ a^2 -2ab+5b^2 -6ac +2bc + 10c^2 \; \; .$

$$ Q^T D Q = H $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ - 1 & 1 & 0 \\ - 3 & - \frac{ 1 }{ 2 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 0 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & - 1 & - 3 \\ 0 & 1 & - \frac{ 1 }{ 2 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 1 & - 1 & - 3 \\ - 1 & 5 & 1 \\ - 3 & 1 & 10 \\ \end{array} \right) $$

$$ E_j^T D_{j-1} E_j = D_j $$ $$ P_{j-1} E_j = P_j $$ $$ E_j^{-1} Q_{j-1} = Q_j $$ $$ P_j Q_j = I $$ $$ P_j^T H P_j = D_j $$ $$ Q_j^T D_j Q_j = H $$

$$ H = \left( \begin{array}{rrr} 1 & - 1 & - 3 \\ - 1 & 5 & 1 \\ - 3 & 1 & 10 \\ \end{array} \right) $$

==============================================

$$ E_{1} = \left( \begin{array}{rrr} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{1} = \left( \begin{array}{rrr} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{1} = \left( \begin{array}{rrr} 1 & - 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{1} = \left( \begin{array}{rrr} 1 & 0 & - 3 \\ 0 & 4 & - 2 \\ - 3 & - 2 & 10 \\ \end{array} \right) $$

==============================================

$$ E_{2} = \left( \begin{array}{rrr} 1 & 0 & 3 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{2} = \left( \begin{array}{rrr} 1 & 1 & 3 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{2} = \left( \begin{array}{rrr} 1 & - 1 & - 3 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{2} = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 4 & - 2 \\ 0 & - 2 & 1 \\ \end{array} \right) $$

==============================================

$$ E_{3} = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & \frac{ 1 }{ 2 } \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{3} = \left( \begin{array}{rrr} 1 & 1 & \frac{ 7 }{ 2 } \\ 0 & 1 & \frac{ 1 }{ 2 } \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{3} = \left( \begin{array}{rrr} 1 & - 1 & - 3 \\ 0 & 1 & - \frac{ 1 }{ 2 } \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{3} = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 0 \\ \end{array} \right) $$

==============================================

$$ P^T H P = D $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ 1 & 1 & 0 \\ \frac{ 7 }{ 2 } & \frac{ 1 }{ 2 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & - 1 & - 3 \\ - 1 & 5 & 1 \\ - 3 & 1 & 10 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 1 & \frac{ 7 }{ 2 } \\ 0 & 1 & \frac{ 1 }{ 2 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 0 \\ \end{array} \right) $$ $$ Q^T D Q = H $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ - 1 & 1 & 0 \\ - 3 & - \frac{ 1 }{ 2 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 0 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & - 1 & - 3 \\ 0 & 1 & - \frac{ 1 }{ 2 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 1 & - 1 & - 3 \\ - 1 & 5 & 1 \\ - 3 & 1 & 10 \\ \end{array} \right) $$

Answer 3

$$ a^2+5b^2+2b=6a+2ab-10 \\ a^2+5b^2+2b-6a-2ab+10=0 \\ \frac{(a+b-4)^2+(a-3b-2)^2}{2}=0 \\ a+b=4,\quad a-3b=2 \\ a=\frac{7}{2},\quad b=\frac{1}{2} $$

The old college course called "Analytic Geometry" used to do this. In the study of conic sections, you would "rotate and translate coordinates". But 50 years ago when "Analytic Geometry" was combined with "Calculus" some of those old classic things were deemed expendable.

Answer 4

Consider the equation as a quadratic in $\,a\,$:

$$ a^2 - 2a(b+3) +5b^2+2b + 10 = 0 $$

Then its reduced discriminant is:

$$ \frac{1}{4} \Delta = (b+3)^2-(5b^2+2b +10) = -4b^2 + 4b -1 = -(2b-1)^2 \;\;\le\;\; 0 $$

It follows that $\,\Delta = 0\,$ for real roots to exist, so $\,b=\dfrac{1}{2}\,$, then substituting back $\,a=b+3=\dfrac{7}{2}\,$.

Answer 5

The equation can be written as $$(2b-1)^2+ a^2+b^2+9-2ab-6a+6b=0$$ $$\Rightarrow (2b-1)^2+ (-a)^2+(b)^2+(3)^2+ (2)(-a)(b)+(2)(3)(-a) +(2)(3)(b)=0$$ $$\Rightarrow (2b-1)^2+(a-b-3)^2=0$$ Hence we get $(a, b) = (\frac {7}{2} , \frac {1}{2})$