Question

quadratic equation has one solution x=6 and x=-8 what is the equation found to be?

Answer 1

You should get your solutions in the form of \(x-root=0\) first of all. In the case of \(x=6\), the new form would be \(x-6=0\). Do the same with the other root (be careful with your signs). You can then multiply these together, like:\[(x-root_1)(x-root_2)\] which you can multiply out to get your quadratic equation in the form \(ax^2+bx+c\)

Answer 2

so the answer would be (x-6)(x-8)

Answer 3

Be careful with your signs. If the root is -8, then when you do \(x-root=0\), you'll get \(x-(-8)=0\). After that, you should expand those brackets.

Answer 4

I should point out that I forgot to write something earlier. The expression should equal 0, so it's \((x-root_1)(x-root_2)=0\)

Answer 5

Then your resulting equation will look like \(ax^2+bx+c=0\)

Answer 6

How do you Inline LaTeX ?

Answer 7

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