Question

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 $(B,D).$

Answer 1

Answer:

A. 33

B. k=32

C. 1

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

E. $B=-2,\ D=5$

Step-by-step explanation:

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

$x_1+x_2=-\dfrac{b}{a},\\ \\x_1\cdot x_2=\dfrac{c}{a},$

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 $-\dfrac{b}{c}=-\dfrac{-4}{4}=1.$

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

$x_1+x_2=\dfrac{1}{3},\\ \\x_1\cdot x_2=-\dfrac{10}{3}.$

Then

$x_1=2,\ x_2=-\dfrac{5}{3}.$

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