Answer:
1. w = -6, -7
2. v = -3, 11
3. Frame's width all around is same, x, and x = 1 inch
4. The other zero of the function is 5 ( x = 5 )
Step-by-step explanation:
1.
For the question given [tex]w^2+13w+42=0[/tex] , we need to find 2 numbers such that their sum is equal to the number before x (which is 13) and their product is equal to the constant (which is 42) AND THEN replace 13w with those 2 numbers(and w).
Such two numbers are 7 and 6
Now we can write:
[tex]w^2+7w+6w+42=0[/tex]
Then we can group the first 2 terms and last 2 terms and take common, then solve.
[tex]w^2+7w+6w+42=0\\w(w+7)+6(w+7)=0\\(w+6)(w+7)=0[/tex]
Since two expressions' product is equal to 0, either w+6 = 0 or w+7 = 0
w+6=0
w = -6
and
w+7 = 0
w= -7
Hence, w = -6, -7
2.
This is similar to Question #1, so we need two numbers that multiplied gives us -33 and added gives us -8.
Such two numbers are -11 and +3
Thus we can replace -8 with this and take common and then solve for v:
[tex]v^2-8v-33=0\\v^2+3v-11v-33=0\\v(v+3)-11(v+3)=0\\(v-11)(v+3)=0[/tex]
So either v-11 =0 or v+3 = 0
v - 11 = 0
v = 11
and
v+3=0
v = -3
Hence, v = -3, 11
3.
Since total area of frame & photo is 120, we need an expression for frame & photo and equate that to 120.
If picture's width is 10 and two sides there is frame of x inches each, so width of whole this is 10 +2x
Similarly, the length is 8 and two sides there is frame of x inches each, so length of whole is 8 +2x
Area of rectangle is length * width
Thus, we have [tex](10+2x)(8+2x)=120[/tex]
We need to find x, so we solve the equation above:
[tex](10+2x)(8+2x)=120\\80+20x+16x+4x^2=120\\4x^2+36x+80-120=0\\4x^2+36x-40=0\\x^2+9x-10=0\\x^2+10x-x-10=0\\x(x+10)-1(x+10)=0\\(x-1)(x+10)=0\\x=1, -10[/tex]
Since length cannot be negative, x = 1 only
4.
Factoring the original equation as we did earlier, we see "what two numbers multiplied gives us -35 and added gives us +2?"
Those 2 numbers are +7 and -5
Thus we can write, group, and solve for x:
[tex]x^2+2x-35=0\\x^2+7x-5x-35=0\\x(x+7)-5(x+7)=0\\(x-5)(x+7)=0[/tex]
So x-5 =0 or x+7 = 0
x+7=0
x = -7 (already given in the problem)
and
x-5=0
x=5
Hence, the other zero of the function is 5
