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# Solving quadratics by taking square roots examples

Sal solves the equation (x+3)²-4=0 and finds the x-intercepts of f(x)=(x-2)²-9.

## Want to join the conversation?

• kindly help i this confusion
(x+7)^2-49=0
(x+7)=+/-7
x=0 or -14
BUT COULDN'T I DO THE FOLLOWING
(x+7)^2-49=0
x^2+14x+49-49=0
x^2+14x=0, then x^2=-14x , divide both sides by x
x=-14
in this i cant discover x=0, how this could happen?! how to follow valid steps and miss one of the solutions ?
• Never divide by the variable. You destroy / lose solutions to the equation.
Instead, use factoring: x^2+14x=0 becomes x(x+14)=0
Use zero product rule: x=0 and x+14=0
Solutions become: x=0 and x=-14

Hope this helps.
• it a parabola is open upwards what does it repersent? (in terms of a throwing arc)
• I just realized that it could represent a ball being thrown into water because it could come back up.
• How are you supposed to solve equations that have square rooted prime numbers?

[Example: x + 8 = ^2]
• You cannot combine numbers and roots, so by subtracting 8, the answer would be x = - 8 + √2. You can get an approximation if you need to, but the exact answer is as above.
• So, would there be an actual way to figure out only one answer.
• you can not figure out only one answer because it has 2 answers.
(1 vote)
• Do you think he started the video a little too late? He doesn't give his usual intro in this video.
• `Yeah. He started the video with, “So pause the video…” so I think he cut off the beginning or something.`
(1 vote)
• how f(x)=(x-2)^2-9 is a function ?
as sal said that functions can only have one output then how can a parabola be a function...?
• A function generates only one output (y-value) for each input value (x-value).
The equation f(x)=(x-2)^2-9 satisfies that requirement. Each time you input a value for "x", you only get one value for "y".

I think you may be confusing "x" and "y". Parabolas will have the same "y" value for 2 different "x" values. That's ok. "y" is not the input.
• how do i add a square root symbol
• On KhanAcademy, if your answer needs to include a square root, then a menu of special symbols is provided when your cursor is in the answer box. If no option is provided, then likely you haven't simplified your answer. For example: sqrt(49) would not be an appropriate answer because it simplifies to 7.

Hope this helps.
• at the problem has (x-2)^2 .... When I tried I thought I couldnt take sqrt of negative numbers?
edit: (Ah it's because the exponent is already there to save them from the sqrt cage)
So it became:
x^2 + 4 = 9
x = sqrt(5)
why I couldnt get the answer this way?
• There are several issues with your approach.

Issue 1) (x-2)^2 does not equal x^2+4. Squaring a binomial always creates a trinomial. (x-2)^2 = x^2-2x-2x+4 = x^2-4x+4

Issue 2) When you applied the square root to: x^2=5, you lost one of the 2 solutions. You should have gotten x=sqrt(5) and x=-sqrt(5).

The correct way to approach the problem is to recognize (x-2)^2 as a perfect square. Thus, sqrt[(x-2)^2] = x-2 (as shown in the video).

Hope this helps.
• Wouldn't it be 'and'? x=-1 and x=-5. My reasoning: It is a parabola and there are two x intercepts.
• I'm not sure what your reasoning mean, since if its an actual reasoning, it will also apply to using 'or'.

x = -1 or x = -5 means the x-value when f(x) = 0. You can't have a point with 2 x-values.
You use x = -1 and x = -5 when you are describing the parabola, so you say
f(x) has 2 x-intercepts, x = -1 and x = -5.