Question

x2-8x=33 using factor formula and quadratic formula. I need help

Answer 1

Hint: x+3 is a factor.

Answer 2

Solve it @Asuarez_7

Answer 3

Thank you, can you help me on a other one?

Answer 4

49^2-1=0

Answer 5

wait,we haven't finish yet..
so what are values for x?

Answer 6

*are the

Answer 7

Is it 3?

Answer 8

Wait no it's x= 11 and x= -3

Answer 9

correct!

Answer 10

One important issue should be mentioned at this point: Just as with linear equations, the solutions to quadratic equations may be verified by plugging them back into the original equation, and making sure that they work, that they result in a true statement. For the above example, we would do the following:\[checking~x=-3~inn~(x+3)(x-11)=0:\]\[([-3]+3)([-3]-11)=0\]\[(0)(-14)=0\]\[0=0\]\[checking~x=11~inn~(x+3)(x-11)=0:\]\[([11]+3)([11]-11)=0\]\[(14)(0)=0\]\[0=0\]So both solutions "check" and are thus verified as being correct.

Answer 11

That was factorisation method,now let's use the quadratic formula to find the values for x

Answer 12

Solving using Quadratic Formula
\[x ^{2}-8x=33 \]
\[Ax ^{2}+Bx+C=0\]
\[x^{2}-8x-33=0\]
a=1 b=-8 c=-33
\[x=\frac{ -b \pm \sqrt{b ^{2}-4ac} }{ 2a }\]
\[x=\frac{ -(-8) \pm \sqrt{8^{2}-4(1)(-33)} }{ 2(1) }\]
\[x=\frac{ 8 \pm \sqrt{64+132} }{ 2 }\]
\[x=\frac{ 8 \pm \sqrt{196} }{ 2 }\]
\[x=\frac{ 8 \pm 14 }{ 2 }\]
\[x _{1}=\frac{ 8 + 14 }{ 2 }\]
\[x _{1}=\frac{ 22 }{ 2 } =11\]
\[x _{2}=\frac{ 8 - 14 }{ 2 }\]
\[x _{2}=-\frac{ 6 }{ 2 } = -3\]

Hope this helped..

Answer 13

Thanks it did but can you pls help me on this One using factoring and quadratic formula 49x^2-1=0

Answer 14

Factorisation:
Difference of two squares:\[x^2-a^2=(x-a)(x+a)\]\[49x^2-1=0\]\[7^2x^2-1^1=0\]\[(7x-1)(7x+1)\]\[7x-1=0,7x+1=0\]\[7x=1,7x=-1\]

Answer 15

Then divide both sides by 7

Answer 16

\[\frac{ 7x }{ 7 }=\frac{ 1 }{ 7 }\]\[\frac{ 7x }{ 7 }=-\frac{ 1 }{ 7 }\]

Answer 17

so what are the values of x? @Asuarez_7

Answer 18

Is it 7?

Answer 19

no it is supposed to be\[x=\frac{ 1 }{ 7 }~and~x=-\frac{ 1 }{ 7 }\]

Answer 20

One important issue should be mentioned at this point: Just as with linear equations, the solutions to quadratic equations may be verified by plugging them back into the original equation, and making sure that they work, that they result in a true statement. For the above example, we would do the following:\[checking~x=\frac{ 1 }{ 7 }~inn~(7x-1)(7x+1)=0:\]\[(7[\frac{ 1 }{ 7 }]-1)(7[\frac{ 1 }{ 7 }]+1)=0\]\[(0)(2)=0\]\[0=0\]\[checking~x=-\frac{ 1 }{ 7 }~inn~(7x-1)(7x+1)=0:\]\[(7[-\frac{ 1 }{ 7 }]-1)(7[-\frac{ 1 }{ 7 }]+1)=0\]\[(-2)(0)=0\]\[0=0\]So both solutions "check" and are thus verified as being correct.