Erwus

The side lengths of the original square and the new square are 3 and 5. Therefore, there are two values of k: 5/3 and -5/3. Note: you can dilate by a negative scale factor, think about the square on a cartesian plane. Therefore sum of the values of k are 0.

Combining like terms, we get the equation:

\(5x^2+4x-6=0\)

Vieta's formulas states that the sum of roots for a quadratic ax^2+bx+c is -b/a. Therefore, the sum of the roots for this equation is -4/5. Since the prom is asking for a+b-8, the answer is -44/5.

A quadratic whose roots are 1 and -1 in factored form looks like this:

\((x-1)(x+1)\)

Expanding this out, and it becomes \(x^2+1\)

Now we fill in the blanks; the first blank is 0, since there is no "x" term. The second blank is 1.

If all three resulting pieces are shorter than 2 units, then the sum of the 3 pieces would be less than six, the length of our big stick. Therefore, there are no ways to make 3 pieces shorter than 2 units, so the probability is 0.

  If we factor the left side of the second equation, we get the equation:

\((x+y)(x^2+xy+y^2)=162+x^2+y^2\)

Since the first equation gives x+y=10, the second equation now becomes:

\(10(x^2+xy+y^2)=162+x^2+y^2 \)

\(10x^2+10xy+10y^2= 162+x^2+y^2\)

\(9x^2+9y^2+10xy=162\)

Now substitute the first equation into the second

\(9(10-y)^2+9y^2+10(10-y)y=162\)

\(9(100-20y+y^2)+9y^2+100y-10y^2=162\)

\(900-180y+9y^2+9y^2+100y-10y^2=162\)

\(8y^2-80y+900=162\)

\(8y^2-80y+738=0\)

However, putting this into the quadratic formula, we find that the discriminant is negative, so therefore there are no real solutions.

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