Melanie made a circular necklace. She puts beads evenly spaced around the necklace. If the 4th bead is directly across from the
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Answer:
Quadratic Equation | Factoring
Solve each quadratic equation using factoring.
1) v² + 5v + 6 = 0
doing middle term factorisation
v²+(3+2)v+6=0
v²+3v+2v+6=0
v(v+3)+2(v+3)=0
(v+3)(v+2)=0
either
<u>v=-3</u>
<u>or</u>
<u>v=-2</u>
2) g² - 3g = 4
keeping all terms in one side
g²-3g-4=0
doing middle term factorisation
g²-(4-1)g-4=0
g²-4g+g-4=0
g(g-4)+1(g-4)=0
(g-4)(g+1)=0
either
<u>g=4</u>
<u>or</u>
<u>g=-1</u>
3)w² + 4w = 0
w(w+4)=0
either
<u>w=0</u>
<u>or</u>
<u>w=-4</u>
4) s² - 8s + 12 = 0
doing middle term factorisation
s²-(6+2)+12=0
s²-6s-2s+12=0
s(s-6)-2(s-6)=0
(s-6)(s-2)=0
either
<u>s=6</u>
<u>or</u>
<u>s=2</u>
5) x ²+ 2x - 35 = 0
doing middle term factorisation
x²+(7-5)x-35=0
x²+7x-5x-35=0
x(x+7)-5(x+7)=0
(x+7)(x-5)=0
either
<u>x=-7</u>
<u>or</u>
<u>x=5</u>
6) r(r + 2) = 99
opening bracket
r²+2r=99
keeping all terms in one side
r²+2r-99=0
r²+(11-9)r-99=0
r²+11r-9r-99=0
r(r+11)-9(r+11)=0
(r+11)(r-9)=0
either
<u>r=-11</u>
<u>or</u>
<u>r=9</u>
7)k(k-4)=-3
opening bracket
k²-4k=-3
keeping all terms in one side
k²-4k+3=0
k²-(3+1)k+3=0
k²-3k-k+3=0
k(k-3)-1(k-3)=0
(k-3)(k-1)=0
either
k=3
or
k=1
8)t²+ 3t + 2 = 0
doing middle term factorisation
t²+(2+1)t+2=0
t²+2t+t+2=0
t(t+2)+1(t+2)=0
(t+2)(t+1)=0
either
<u>t</u><u>=</u><u>-</u><u>2</u>
<u>or</u>
<u>t</u><u>=</u><u>-</u><u>1</u>
9)m ^ 2 - 81 = 0
m²=81
doing square root in both side
<u>m=±9</u>
<u>either</u>
<u>m</u><u>=</u><u>9</u>
<u>or</u>
<u>m</u><u>=</u><u>-</u><u>9</u>
10) h²- 17h + 70 = 0
doing middle term factorisation
h²-(10+7)h+70=0
h²-10h-7h+70=0
h(h-10)-7(h-10)=0
(h-10)(h-7)=0
either
<u>h</u><u>=</u><u>1</u><u>0</u>
<u>or</u>
<u>h</u><u>=</u><u>7</u>