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A. What is the sum of the squares of the roots of $x^2 - 5x - 4 = 0$?

B. One root of $x^2 + 12x + k = 0$ is twice the other root. Find $k.$
C. What is the sum of the roots of the quadratic $4x^2 - 4x - 4$?
D. Jimmy is trying to factor the quadratic equation $ax^2 + bx + c = 0.$ He assumes that it will factor in the form
\[ax^2 + bx + c = (Ax + B)(Cx + D),\]where $A,$ $B,$ $C,$ and $D$ are integers. If $a = 4,$ and Jimmy wants to find the value of $A,$ what are the possible values he should check, in order to find $A$?
E. Brandy is trying to factor the quadratic $3x^2 - x - 10.$ She starts by assuming that the quadratic factors as
\[3x^2 - x - 10 = (x + B)(3x + D),\]for some integers $B$ and $D.$ After some work, Brandy successfully factors the quadratic. Find the ordered pair $(B,D).$

User Zolio
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1 Answer

6 votes

Answer:

A. 33

B. k=32

C. 1

D.
\pm 1,\ \pm 2,\ \pm 4

E.
B=-2,\ D=5

Explanation:

In all parts for the quadratic equation
ax^2+bx+c=0 use Vieta's formulas


x_1+x_2=-(b)/(a),\\ \\x_1\cdot x_2=(c)/(a),

where
x_1,\ x_2 are the roots of the quadratic equation.

A. For the equation
x^2-5x-4=0,


x_1+x_2=5,\\ \\x_1\cdot x_2=-4.

Then


(x_1+x_2)^2=x_1^2+2x_1\cdot x_2+x_2^2,\\ \\5^2=x_1^2+x_2^2+2\cdot (-4),\\ \\x_1^2+x_2^2=25+8=33.

B. One of the roots of
x^2+12x+k=0 is twice the other root, then
x_2=2x_1. By the Vieta's formulas,


x_1+x_2=3x_1=-12,\\ \\x_1\cdot x_2=2x_1^2=k.

Then
x_1=-4 and
k=2x_1^2=2\cdot (-4)^2=2\cdot 16=32.

C. The sum of the roots of the quadratic
4x^2-4x-4 is
-(b)/(c)=-(-4)/(4)=1.

D. Note that


(Ax+B)(Cx+D)=ACx^2+x(AD+BC)+BD,

then
AC=a=4. If
A,\ B,\ C,\ D are integers, then you should check
A=\pm 1,\ \pm 2,\ \pm 4.

E. Consider
3x^2 - x - 10 = (x + B)(3x + D). Note that


x_1+x_2=(1)/(3),\\ \\x_1\cdot x_2=-(10)/(3).

Then


x_1=2,\ x_2=-(5)/(3).

Then
3x^2 - x - 10 = (x -2)(3x+5), hence
B=-2,\ D=5.


User Saki
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