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## Algebra 1

### Course: Algebra 1 > Unit 14

Lesson 3: Solving by taking the square root- Solving quadratics by taking square roots
- Solving quadratics by taking square roots
- Quadratics by taking square roots (intro)
- Solving quadratics by taking square roots examples
- Quadratics by taking square roots
- Quadratics by taking square roots: strategy
- Quadratics by taking square roots: strategy
- Solving quadratics by taking square roots: with steps
- Quadratics by taking square roots: with steps
- Solving simple quadratics review

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# Solving quadratics by taking square roots: with steps

CCSS.Math: ,

Sal discusses the exact order of steps in the process of solving the equation 3(x+6)^2=75. Created by Sal Khan.

## Want to join the conversation?

- Do we have to write the plus or minus signs?(9 votes)
- Miles,

When you take the square root of both sides of an equation, you get two answers.

In the video, Sal had

(x+6)²=25 Take the square root of both sides

x+6 = ±√25

x+6=±5

The plus or minus sign means you have two equations

x+6 = 5 and x+6 = -5

You don't have to write the ± sign but if you don't, you then need to write out both equations x+6 = 5 and x+6 = -5 and then solve each equation to get your two answers.

The ± sign is used to indicate that you have two equations you are solving at the same time.

I hope that helps make it click for you.(25 votes)

- how to find roots of a quadratic equation if b^2-4ac < 0 , if b^2-4ac is a perfect square i.e. (-64) .(8 votes)
- Undefined. Undefined because sqrt(<0) is undefined.(5 votes)

- It seems that the exercise for this video is missing. Can someone tell me where these exercises are located?(5 votes)
- It would be in "Algebra">"Quadratic Equations">"Solving Quadratics by Taking Square Root" and all the way at the bottom is "Understanding the Equation Solving Process" where a problem like this should appear.(9 votes)

- At3:10, why is it -6 plus or minus 5 instead of 5 plus or minus -6?(5 votes)
- You started with
`x+6=±5`

, so subtracting 6 from both sides yields`x=-6±5`

.

If you started with`x+6=+5 or x+6=-5`

and subtracted 6 from both sides, you'd get`x=-6+5 or x=-6-5`

. And you could recombine those last two possibility back to just`x=-6±5`

. So it's the same thing either way: the`±`

sign is attached to the 5.(5 votes)

- Is it okay if we just put the steps and ignore figuring out the exact answer?(2 votes)
- Yes. Sal figured out the exact answer just for extra instruction.(7 votes)

- on the options shown on the question,2:00Sal chooses a "take the square root of both sides" option. not to make myself dumb if it's too obvious, but i fail to see the difference between "square both sides" and "take the square root of both sides". HEELLP!(4 votes)
- They are opposite operations.

If you square 4, then you are doing 4^2 = 4*4 = 16

If you take the square root of 4, then you are doing sqrt(4) = 2

Do you see the difference?

If Sal had squared both sides of: (x+6)^2=25, here's what would have happened:

[(x+6)^2]^2 = 25^2

(x+6)^4 = 625

Squaring both sides just makes the equation even more complicated.

Hope this helps.(2 votes)

- I understand the reason Sal is putting +/- in front of the square rooted number. However why is it that we do this only in quadratics and not other equations?(3 votes)
- Quadratics are not the only place where the +/- is used. We may be using functions like absolute value. But of the not-so-many places in math where the +/- will get used, quadratics is one of the places where it will be appearing most with little to no absence. We need it for absolute value so we can find which x-values, the positive and negative, will be adequate values for the overall function (and have outputs to the same p-value).

Most other places, however, do not need to use the +/- as a way to find the equation(s) and may only have at most one possible solution, not two.(2 votes)

- How would you solve:

h(x) = -3(x-2)(x+2)

I am found the vertices x=2 and x= -2, but I am not sure how to find the axis of symmetry, AKA the maximum or minimum. PLEASE HELP!!(3 votes)- to find the x-coordinate of the vertex (which is also the axis of symmetry):

1. first find the 2 x-intercepts(when y=0)-> (x[1],0) (x[2],0)

2.add the x values of the x-intercepts then divide them by 2 -> (x[1]+x[2])/2

In your question the x-intercepts are (2,0)(-2,0)

(2+-2)/2 = 0/2 = 0

so the x-coordinate of the vertex and the axis of symmetry is 0

To find the max/min value we need to find the y- coordinate of the vertex

to do that we substitute the x coordinate of the vertex in the equation

-> h(0)=-3(0-2)(0+2)

h(0)= -3(-2)(2)

h(0)= -3(-4)

h(0)= -12

so the vertex is (0,12)

and the max value(it has a maximum value because it opens down) is 12(1 vote)

- so what if I have a problem like x=(x -7)2-4 ? (I can't do little two sorry)(1 vote)
- Anthony, I think that Rhyann97 meant (x-7)^2, since he/she said something about "little two".

x=(x-7)^2-4

=> x=x^2-14x+49-4

=> x^2-15x+45=0

=> Quadratic formula: (15+ - sqrt(225-180))/2

=> (15+ - sqrt(45))/2

=> (15+ - 3sqrt(5))/2.(5 votes)

- how can you solve:

1/2(x-1)^2 +5=23 can we solve it using this method(1 vote)- Yes... you can.

1) subtract 5 from both sides: 1/2 (x-1)^2 = 18

2) Then multiply both sides by 2 to eliminate the fraction: (x-1)^2 = 36

3) Take square root of both sides: x - 1 = +/- sqrt(36) which simplifies to x - 1 = +/- 6

4) Add 1 to both sides: x = 1 +/- 6

So x = 7 and x = -5(2 votes)

## Video transcript

Use the cards below to create
a list of steps in order that will solve the
following equation. 3 times x plus 6
squared is equal to 75. And I encourage you to
pause this video now and try to figure it out on your own. Figure out which of these
steps and in what order you would do to
solve for x here. So I'm assuming
you've given it a go. So let's try to work
through it together. And first, let me just
rewrite the equation. So we have 3 times the
quantity x plus 6 squared is equal to 75. So what I want to do is I want
to isolate the x plus 6 squared on the left-hand side. Or another way of
thinking about it-- I don't want this
3 here anymore. So how would I
get rid of that 3? Well, I could divide
the left-hand side by 3. But if I do that to only
one side of the equation, it won't be equal anymore. These two things in yellow
were equal to each other. If I want the equalities
to hold, anything that I do to the
left-hand side, I have to do the right-hand side. So let me divide
that by 3 as well. And so on the
left-hand side, I am left with x plus 6 squared
is equal to 75 divided by 3. So 75 divided by 3 is 25. So actually, let me just
pick out the first one I did. I divided both sides by 3. So that was my first step then. Let me write that
in a darker color. So that was my first
step right over there. Now let's think about
what we're doing. We're saying that something
squared is equal to 25. So this something could be
the positive or negative square root of 25. So we could write
this as x plus 6 is equal to the plus or
minus square root of 25. So I'm essentially taking
the positive and negative square root of both sides. So, let's see. This looks like this step. I took the square
root of both sides. That's step number two. And so, let me
just rewrite this. This is the same thing as x plus
6 is equal to plus or minus 5. And now I want to just have
an x on the left-hand side. I want to solve for x. That's the goal
from the beginning. So I would like to
get rid of this 6. Well, the easiest
way to do that is to subtract 6 from
the left-hand side. But just like
before, I can't just do it from one side
of an equation. Then the equality
wouldn't be true. We're literally
saying that x plus 6 is equal to plus or minus 5. So x plus 6 minus
6 is going to be equal to plus or
minus 5 minus 6. Or actually, let me
write it this way. So let me subtract
6 from both sides. On the left-hand side,
I'm left with an x. And on the right-hand side,
I could write it this way. Let me do it in
that green color. I have negative 6
plus or minus 5. So what are the
possible values of x? Or actually, I keep forgetting. We don't have to actually
give the value for x. We just have to say
what steps we did. So then, let's see. After we took the square
root of both sides, we then subtracted
6 from both sides. So that was step three
right over there. Then that got us to essentially
the two possible x's that would satisfy this
equation right over here. And just for fun, let's
actually solve it all the way. So if we solve it all the way,
so x is equal to negative 6 plus 5 is negative 1, or x is
equal to negative 6 minus 5 is negative 11. And you could verify
that both of these work. If you put either of them in
here-- if you put negative 1 here, you get negative 1
plus 6 squared is 5 squared. If you put negative 11 here,
it's negative 11 plus 6 is negative 5 squared. Obviously either plus or minus
5 squared is going to be 25. 25 times 3 is 75. So these are our three steps. We divided both sides by 3. Then we took the square
root of both sides. Then we subtracted
6 from both sides. And then we were
essentially done. So let's input those steps. So the first thing we did,
we divide both sides by 3. That's the first thing we did. And then we took the
square root of both sides. And then we subtracted
6 from both sides. We got it right.