Question

Find all pairs of real numbers (a,b) such that (x-a)^2+(2x-b)^2=(x-3)^2+(2x)^2

Answer 1

Answer:

$(3,0)$

$(\frac{-9}{5},\frac{12}{5})$

Step-by-step explanation:

Let's expand both sides.

I'm going to use the following identity to expand the binomial squared expressions: $(u+v)^2=u^2+2uv+v^2$ or $(u-v)^2=u^2-2uv+v^2$.

Left-hand side:

$(x-a)^2+(2x-b)^2$

$(x^2-2ax+a^2)+((2x)^2-2b(2x)+b^2)$

$x^2-2ax+a^2+4x^2-4bx+b^2$

Reorder so $x^2$'s are together and that $x[tex]'s are together.[tex](x^2+4x^2)+(-2ax-4bx)+(a^2+b^2)$

$5x^2+(-2a-4b)x+(a^2+b^2)$

Right-hand side:

$(x-3)^2+(2x)^2$

$x^2-2(3)x+9+4x^2$

$x^2-6x+9+4x^2$

Reorder so $x^2$'s are together and that $x[tex]'s are together.[tex](x^2+4x^2)+(-6x)+9$

$5x^2-6x+9$

Now let's compare both sides.

If we want both sides to appear exactly the same we need to choose values $a$ and $b$ such the following are true equations:

$-2a-4b=-6$

$a^2+b^2=9$

So if we solve the system we can find the values $a$ and $b$ such that the left=right.

Let's solve the first equation for $a$ in terms of $b$.

Add $2a$ on both sides:

$-4b=-6+2a$

Divide both sides by -4:

$b=\frac{-6+2a}{-4}$

Reduce (divide top and bottom by -2):

$b=\frac{3-a}{2}$

Now let's plug this into second equation:

$a^2+b^2=9$

$a^2+(\frac{3-a}{2})^2=9$

$a^2+\frac{9-6a+a^2}{4}=9$ (I used the identity $(u-v)^2=u^2-2uv+v^2$)

Multiply both sides by 4 to clear the fractions from the problem:

$4a^2+(9-6a+a^2)=36$

Combine like terms on left hand side:

$4a^2+a^2-6a+9=36$

$5a^2-6a+9=36$

Subtract 36 on both sides:

$5a^2-6a-27=0$

Now let's try to factor.

We are going to try to find two numbers that multiply to be 5(-27) and add to be -6.

5(-27)=(5*3)(-9)=15(-9)=-15(9) while -15+9=-6.

So let's replace $-6a$ with $-15a+9a$ and factor by grouping.

$5a^2-15a+9a-27=0$

$5a(a-3)+9(a-3)=0$

$(a-3)(5a+9)=0$

This implies $a-3=0$ or $5a+9=0$.

Solving the first is easy. Just ad 3 on both sides to get: $a=3$.

The second requires two steps. Subtract 9 and then divide by 5 on both sides.

$5a=-9$

$a=\frac{-9}{5}$.

So let's go back to finding $b$ now that we know the $a$ values.

If $a=3$ and $b=\frac{3-a}{2}$,

then $b=\frac{3-3}{2}=0$.

So one ordered pair $(a,b)$ that satisfies the equation is:

$(3,0)$.

If $a=\frac{-9}{5}$ and $b=\frac{3-a}{2}$,

then $b=\frac{3-\frac{-9}{5}}{2}$.

Let's multiply top and bottom by 5 to clear the mini-fraction.

$b=\frac{15-(-9)}{10}$

$b=\frac{24}{10}$

$b=\frac{12}{5}$

So one ordered pair $(a,b)$ that satisfies the equation is:

$(\frac{-9}{5},\frac{12}{5})$.