Question

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Answer 1

You can solve this by completing the square or using the quadratic formula.

Quadratic formula:
For a polynomial \(ax^2+bx+c=0, a\ne0\), the solutions are given by
\[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]
You have \(a=1, b=6,c=7\) so the solutions are
\[x=\frac{-6\pm\sqrt{6^2-4(1)(7)}}{2(1)}\]

Answer 2

Completing the square:

Rearrange your equation in the form \(ax^2+bx = c\), then divide the entire equation by \(a\) if \(a\ne 1\)

From here on, I'll assume the equation looks like \(x^2+Bx = C\)
We want to write the left hand side as a perfect square of the form \((x+p)^2\). After doing that, we can take the square root of both sides and find the values of \(x\) that are solutions.

If we multiply out \[(x+p)^2 = (x+p)(x+p) = x^2+2px + p^2\]that gives us a template for what we need to add to the left hand side to make it a perfect square. (We'll also add the same quantity to the right hand side to keep it equal, very important!) We can that if we take half of the coefficient of the \(x\) term, square it, and add it to what we have, we'll have something we can write as a perfect square. Let's try it on our equation:

\[x^2+6x +7= 0\]\[x^2+6x=-7\]Take half of 6, square it, and add to both sides:
\[x^2+6x + (6/2)^2 = -7 + (6/2)^2\]\[x^2+6x + 9 = 2\]Rewrite the left hand side as a perfect square
\[(x+3)^2 = 2\]Take square roots of both sides
\[x+3 = \pm\sqrt{2}\]We have to take both the positive and negative square root values of the right hand side to get the two solutions.
Now solve that for \(x\) and you're all done.

Answer 3

oh my goodness..

Answer 4

trinomials have always have two answers

Answer 5

Wait a minute, where did the "all over 2" come from?

Answer 6

No i mean how did you solve for it to where it became a fraction?

Answer 7

@raaachel no, trinomials do not always have two answers. consider \[x^4+2x+1\]which has 4 solutions:

\[\{\{x\to -1.\},\{x\to -0.543689\},\{x\to 0.771845\, +1.11514 i\},\{x\to 0.771845\, -1.11514 i\}\}\]

Quadratics, which are often trinomials always have 2 solutions. In general, you have as many solutions as the highest exponent in the polynomial.

Answer 8

@malehuman01 I just showed you how to do it. All you have to do is fill in the blanks in the quadratic formula. I even did that, less the simplification.

Answer 9

you have \(x+3 = \pm\sqrt{2}\) and you want to solve for x.

Subtract -3 from both sides \[\large x-3+3 = -3 \pm \sqrt{2}\]Simplify

Answer 10

I meant quadratic trinomials, we just learnt that they will have two answers

Answer 11

Any quadratic, trinomial or not, will have 2 solutions. The trinomial part is a bit of a red herring.

Answer 12

Just showed ya.

Answer 13

\[x=\frac{-6\pm\sqrt{6^2-4(1)(7)}}{2(1)}\]
Just evaluate that expression.

Answer 14

\[x=\frac{-6\pm\sqrt{36-28}}{2}\]

Answer 15

That's shorthand for \[x=\frac{-6+\sqrt{6^2-4(1)(7)}}{2(1)}\]and\[x=\frac{-6-\sqrt{6^2-4(1)(7)}}{2(1)}\]

Answer 16

yep

Answer 17

indeed. \[\sqrt{8} = \sqrt{2*2*2} = \sqrt{2*2}*\sqrt{2} = 2\sqrt{2}\]