Respuesta :
Answer:
[tex](3,0)[/tex]
[tex](\frac{-9}{5},\frac{12}{5})[/tex]
Step-by-step explanation:
Let's expand both sides.
I'm going to use the following identity to expand the binomial squared expressions: [tex](u+v)^2=u^2+2uv+v^2[/tex] or [tex](u-v)^2=u^2-2uv+v^2[/tex].
Left-hand side:
[tex](x-a)^2+(2x-b)^2[/tex]
[tex](x^2-2ax+a^2)+((2x)^2-2b(2x)+b^2)[/tex]
[tex]x^2-2ax+a^2+4x^2-4bx+b^2[/tex]
Reorder so [tex]x^2[/tex]'s are together and that [tex]x[tex]'s are together.
[tex](x^2+4x^2)+(-2ax-4bx)+(a^2+b^2)[/tex]
[tex]5x^2+(-2a-4b)x+(a^2+b^2)[/tex]
Right-hand side:
[tex](x-3)^2+(2x)^2[/tex]
[tex]x^2-2(3)x+9+4x^2[/tex]
[tex]x^2-6x+9+4x^2[/tex]
Reorder so [tex]x^2[/tex]'s are together and that [tex]x[tex]'s are together.
[tex](x^2+4x^2)+(-6x)+9[/tex]
[tex]5x^2-6x+9[/tex]
Now let's compare both sides.
If we want both sides to appear exactly the same we need to choose values [tex]a[/tex] and [tex]b[/tex] such the following are true equations:
[tex]-2a-4b=-6[/tex]
[tex]a^2+b^2=9[/tex]
So if we solve the system we can find the values [tex]a[/tex] and [tex]b[/tex] such that the left=right.
Let's solve the first equation for [tex]a[/tex] in terms of [tex]b[/tex].
Add [tex]2a[/tex] on both sides:
[tex]-4b=-6+2a[/tex]
Divide both sides by -4:
[tex]b=\frac{-6+2a}{-4}[/tex]
Reduce (divide top and bottom by -2):
[tex]b=\frac{3-a}{2}[/tex]
Now let's plug this into second equation:
[tex]a^2+b^2=9[/tex]
[tex]a^2+(\frac{3-a}{2})^2=9[/tex]
[tex]a^2+\frac{9-6a+a^2}{4}=9[/tex] (I used the identity [tex](u-v)^2=u^2-2uv+v^2[/tex])
Multiply both sides by 4 to clear the fractions from the problem:
[tex]4a^2+(9-6a+a^2)=36[/tex]
Combine like terms on left hand side:
[tex]4a^2+a^2-6a+9=36[/tex]
[tex]5a^2-6a+9=36[/tex]
Subtract 36 on both sides:
[tex]5a^2-6a-27=0[/tex]
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 [tex]-6a[/tex] with [tex]-15a+9a[/tex] and factor by grouping.
[tex]5a^2-15a+9a-27=0[/tex]
[tex]5a(a-3)+9(a-3)=0[/tex]
[tex](a-3)(5a+9)=0[/tex]
This implies [tex]a-3=0[/tex] or [tex]5a+9=0[/tex].
Solving the first is easy. Just ad 3 on both sides to get: [tex]a=3[/tex].
The second requires two steps. Subtract 9 and then divide by 5 on both sides.
[tex]5a=-9[/tex]
[tex]a=\frac{-9}{5}[/tex].
So let's go back to finding [tex]b[/tex] now that we know the [tex]a[/tex] values.
If [tex]a=3[/tex] and [tex]b=\frac{3-a}{2}[/tex],
then [tex]b=\frac{3-3}{2}=0[/tex].
So one ordered pair [tex](a,b)[/tex] that satisfies the equation is:
[tex](3,0)[/tex].
If [tex]a=\frac{-9}{5}[/tex] and [tex]b=\frac{3-a}{2}[/tex],
then [tex]b=\frac{3-\frac{-9}{5}}{2}[/tex].
Let's multiply top and bottom by 5 to clear the mini-fraction.
[tex]b=\frac{15-(-9)}{10}[/tex]
[tex]b=\frac{24}{10}[/tex]
[tex]b=\frac{12}{5}[/tex]
So one ordered pair [tex](a,b)[/tex] that satisfies the equation is:
[tex](\frac{-9}{5},\frac{12}{5})[/tex].
