How to solve y = x^2 - 2x - 6 y = 4x + 10
I know that I should make them equal to each other: x^2 - 2x - 6 = 4x + 10
correct and then combine like terms (set the equation equal to zero) as you can see highest exponent is 2 so it's a quadratic equation
x^2 - 6x + 4 = 0 ?
x^2 - 6x + 4 = 0 x^2 - 6x + 4 - 4 = 0 - 4 x^2 - 6x + (-6/2)^2 = -4 + (-6/2)^2 x^2 - 6x + 9 = -4 + 9 x^2 - 6x + 9 = 5 (x - 3)^2 = \(\sqrt{5}\)
\[x^2-6x-16=0\]
hmm no its not x^2-6x+4=0 10 is positive at right side how would you move that to the left ? subtract or add ?
\[\large\rm x^2 - 6x \color{reD}{ - 6 }=\color{Red}{+ 10}\]
Oh okay. I wrote the question wrong in my paper. but the question posted here is correct .
x^2 - 6x - 16 = 0
correct now factor it
ok
\((x - 3)^2 - 5\)
sorry \((x-3)^2 = 5\)
\[x^2 - 6x - 16 = 0 \] factor it in order words solve for x you don't need to complete the square use quadratic formula or AC method what two numbers would you multiply to get the product of AC same two numbers should be add up to b (coefficient of x term)
i got (x-8)(x+2)
-8 times 2 = -16 correct -8+2 = -6 good now set both parentheses equal to zero then solve for x
x = 8 x = -2
It's very important that you check you "solutions" by substituting each, one at a time, into the original equations. Do you obtain the same ' y ' value in each case?
looks good now pick your fvt equation (doesn't matter first or 2nd) substitute x for each value to find y
ok
ye. to double check you should use both equation to find y.
so i would do 8^2 - 2(8) - 6 64 - 16 - 6 42
looks good . thanks for postin the work:=))
what would i do with 4x + 10?
@Nnesha
you have 2 x value substitute each x to get the y you can use 2nd equation like i said doesn't matter which one you pick you will get the same answer y=x^2-2x-6 when x = 8 y = ? when x=-2 y = ? OR substitute 8 and -2 into the 2nd equation for x y=4x+10
oh okay. i got it. so i will get two coordinates right?
yes correct.
okay thanks!
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