Question

solve the equation by completing the square:
\[2x^2-10x+3=0\]

Answer 1

ax^2+bx+c=0
The coefficient of x^2 must be 1, so factor 'a' out

a(x^2+bx/a)+c=0
divide bx/a by 2,then set the x^2 becomes x only,eliminate the x from b/a, then square the whole bracket

you ended up with

a(x+b/2a)^2+c=0

then, you need to bring the b/2a OUT the brackets and square it, and when you bring it out, it is always negative, be sure that the 'a' will be multiplying the one you bought it out

a(x+b/2a)^2-(b/2a)^2(a)+c=0

Answer 2

so I take-2 from the a part and fro, the b part to eliminate the X

Answer 3

i got lost some place

Answer 4

x-5+3 \[\sqrt{x-5+3}\]

Answer 5

\[2x^2−10x+3=0\]
\[2(x^2-5x)+3=0\]
\[2(x-5/2)^2-\frac{25}{4}(2)+3=0\]

Answer 6

we wrote it different but x = 2 correct?

Answer 7

\[2(x-5/2)^2-\frac{25}{4}(2)+3=0\]
\[2(x-5/2)^2-\frac{19}{2}=0\]

Answer 8

2-5+3=0

Answer 9

\[\begin{array}{l} 2 \left(x-\frac{5}{2}\right)^2-\frac{19}{2}=0 \\ \text{ } \\ 2 \left(x-\frac{5}{2}\right)^2=\frac{19}{2} \\ \text{} \\ \left(x-\frac{5}{2}\right)^2=\frac{19}{4} \\ \text{} \\ \left|x-\frac{5}{2}\right|=\frac{\sqrt{19}}{2} \\ \text{} \\ x-\frac{5}{2}=-\frac{\sqrt{19}}{2}\text{ or }x-\frac{5}{2}=\frac{\sqrt{19}}{2} \\ \text{}\text{} \\ x=\frac{1}{2} \left(5-\sqrt{19}\right)\text{ or }x-\frac{5}{2}=\frac{\sqrt{19}}{2} \\ \text{ }\text{ } \\ x=\frac{1}{2} \left(5-\sqrt{19}\right)\text{ or }x=\frac{1}{2} \left(5+\sqrt{19}\right) \\\end{array}\]

Answer 10

i just got the -19/2 part when i was re-doing

Answer 11

k i think (hope) I followed, my steps were not as neat but i re-did them

Answer 12

ok

Answer 13

you are a lifesaver, i was on the wrong path big time at first