we'll start off by grouping some
![\bf 3x^2-3x-5=13\implies (3x^2-3x)-5=13\implies 3(x^2-x)-5=13 \\\\\\ 3(x^2-x)=18\implies (x^2-x)=\cfrac{18}{3}\implies (x^2-x)=6\implies (x^2-x+~?^2)=6](https://tex.z-dn.net/?f=%5Cbf%203x%5E2-3x-5%3D13%5Cimplies%20%283x%5E2-3x%29-5%3D13%5Cimplies%203%28x%5E2-x%29-5%3D13%20%5C%5C%5C%5C%5C%5C%203%28x%5E2-x%29%3D18%5Cimplies%20%28x%5E2-x%29%3D%5Ccfrac%7B18%7D%7B3%7D%5Cimplies%20%28x%5E2-x%29%3D6%5Cimplies%20%28x%5E2-x%2B~%3F%5E2%29%3D6)
so we have a missing guy at the end in order to get the a perfect square trinomial from that group, hmmm, what is it anyway?
well, let's recall that a perfect square trinomial is
![\bf \qquad \textit{perfect square trinomial} \\\\ (a\pm b)^2\implies a^2\pm \stackrel{\stackrel{\text{\small 2}\cdot \sqrt{\textit{\small a}^2}\cdot \sqrt{\textit{\small b}^2}}{\downarrow }}{2ab} + b^2](https://tex.z-dn.net/?f=%5Cbf%20%5Cqquad%20%5Ctextit%7Bperfect%20square%20trinomial%7D%20%5C%5C%5C%5C%20%28a%5Cpm%20b%29%5E2%5Cimplies%20a%5E2%5Cpm%20%5Cstackrel%7B%5Cstackrel%7B%5Ctext%7B%5Csmall%202%7D%5Ccdot%20%5Csqrt%7B%5Ctextit%7B%5Csmall%20a%7D%5E2%7D%5Ccdot%20%5Csqrt%7B%5Ctextit%7B%5Csmall%20b%7D%5E2%7D%7D%7B%5Cdownarrow%20%7D%7D%7B2ab%7D%20%2B%20b%5E2)
so we know that the middle term in the trinomial, is really 2 times the other two without the exponent, well, in our case, the middle term is just "x", well is really -x, but we'll add the minus later, we only use the positive coefficient and variable, so we'll use "x" to find the last term.
![\bf \stackrel{\textit{middle term}}{2(x)(?)}=\stackrel{\textit{middle term}}{x}\implies ?=\cfrac{x}{2x}\implies ?=\cfrac{1}{2}](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7B%5Ctextit%7Bmiddle%20term%7D%7D%7B2%28x%29%28%3F%29%7D%3D%5Cstackrel%7B%5Ctextit%7Bmiddle%20term%7D%7D%7Bx%7D%5Cimplies%20%3F%3D%5Ccfrac%7Bx%7D%7B2x%7D%5Cimplies%20%3F%3D%5Ccfrac%7B1%7D%7B2%7D)
so, there's our fellow, however, let's recall that all we're doing is borrowing from our very good friend Mr Zero, 0, so if we add (1/2)², we also have to subtract (1/2)²
![\bf \left( x^2 -x +\left[ \cfrac{1}{2} \right]^2-\left[ \cfrac{1}{2} \right]^2 \right)=6\implies \left( x^2 -x +\left[ \cfrac{1}{2} \right]^2 \right)-\left[ \cfrac{1}{2} \right]^2=6 \\\\\\ \left(x-\cfrac{1}{2} \right)^2=6+\cfrac{1}{4}\implies \left(x-\cfrac{1}{2} \right)^2=\cfrac{25}{4}\implies x-\cfrac{1}{2}=\sqrt{\cfrac{25}{4}} \\\\\\ x-\cfrac{1}{2}=\cfrac{\sqrt{25}}{\sqrt{4}}\implies x-\cfrac{1}{2}=\cfrac{5}{2}\implies x=\cfrac{5}{2}+\cfrac{1}{2}\implies x=\cfrac{6}{2}\implies \boxed{x=3}](https://tex.z-dn.net/?f=%5Cbf%20%5Cleft%28%20x%5E2%20-x%20%2B%5Cleft%5B%20%5Ccfrac%7B1%7D%7B2%7D%20%5Cright%5D%5E2-%5Cleft%5B%20%5Ccfrac%7B1%7D%7B2%7D%20%5Cright%5D%5E2%20%5Cright%29%3D6%5Cimplies%20%5Cleft%28%20x%5E2%20-x%20%2B%5Cleft%5B%20%5Ccfrac%7B1%7D%7B2%7D%20%5Cright%5D%5E2%20%5Cright%29-%5Cleft%5B%20%5Ccfrac%7B1%7D%7B2%7D%20%5Cright%5D%5E2%3D6%20%5C%5C%5C%5C%5C%5C%20%5Cleft%28x-%5Ccfrac%7B1%7D%7B2%7D%20%5Cright%29%5E2%3D6%2B%5Ccfrac%7B1%7D%7B4%7D%5Cimplies%20%5Cleft%28x-%5Ccfrac%7B1%7D%7B2%7D%20%5Cright%29%5E2%3D%5Ccfrac%7B25%7D%7B4%7D%5Cimplies%20x-%5Ccfrac%7B1%7D%7B2%7D%3D%5Csqrt%7B%5Ccfrac%7B25%7D%7B4%7D%7D%20%5C%5C%5C%5C%5C%5C%20x-%5Ccfrac%7B1%7D%7B2%7D%3D%5Ccfrac%7B%5Csqrt%7B25%7D%7D%7B%5Csqrt%7B4%7D%7D%5Cimplies%20x-%5Ccfrac%7B1%7D%7B2%7D%3D%5Ccfrac%7B5%7D%7B2%7D%5Cimplies%20x%3D%5Ccfrac%7B5%7D%7B2%7D%2B%5Ccfrac%7B1%7D%7B2%7D%5Cimplies%20x%3D%5Ccfrac%7B6%7D%7B2%7D%5Cimplies%20%5Cboxed%7Bx%3D3%7D)
Steps?
A graph shows zeros to be ±3. Factoring those out leaves the quadratic
(x-2)² +1
which has complex roots 2±i.
The function has roots -3, 3, 2-i, 2+i.
Answer:
The inverse is 1/3 (x+1)
Step-by-step explanation:
y = 3x-1
Exchange x and y
x = 3y-1
Solve for y
Add 1 to each side
x+1 = 3y-1+1
x+1 = 3y
Divide each side by 3
(x+1)/3 = 3y/3
1/3(x+1) =y
The inverse is 1/3 (x+1)
Answer:
Step-by-step explanation:
1. p║ q
50+130 = 180
If the same side interior angles are supplementary angles then the lines are parallel.
2. p║ q
70 = 70
If the corresponding angles are congruent the lines are parallels.
4. p║ q
x = x
If alternating exterior angles are congruent then the lines are parallel.
5. we do not know if p is parallel with q
We have given that 2 vertical angles are congruent yet that is not enough to tell us about the relation between the 2 lines.
7. For the lines p and q to be parallel we need the corresponding angles 3x and 45 to be congruent so therefore equal in measure.
3x= 45 , divide both sides by 3
x= 15
For x = 15 the p║ q
8. For the lines p and q to be parallel we need the corresponding angles 120 and (2x+10) to be congruent so therefore equal in measure.
2x+10 = 120, subtract 10 from both sides
2x = 110, divide both sides by 2
x = 55
For x = 55 the p║ q