Question

Solve the equation by finding the square roots.

1. 4x^2 -4x+1=36

2. x^2 -2/3x+1/9=1

3. 9x^2 +12x +4 =5

Answer 1

1. \(4x^2 -4x+1=36\)

First, can you factor the left side?
Hint: It's the square of a binomial.

Answer 2

wait is it x^2 -x +1/4 =9??

Answer 3

No. Just take \(4x^2 - 4x + 1 \) and factor it into the square of a binomial.

Answer 4

What you wrote is correct, but it doesn't help.

Answer 5

i dont get it what do u mean by square binomia.ll

is it 4x^ -2x +1?

Answer 6

Haven't you learned how to factor trinomials?
For example,
\(x^2 + 3x + 2\)
factors into \( (x + 2)(x + 1) \).

Answer 7

but i dk how to factor this one cuz it has a coefficient infront of the x^

Answer 8

can u please show me how to do this one....

Answer 9

Ok, here it is.
To factor a trinomial of the form ax^2 + bx + c, start by
1. multiplying a and c together.
2. Then find 2 numbers that multiply to ac and add to b. Let's call them m and n.
3. Then break up the middle term of the trinomial into bx = mx + nx.
4. Then factor the polynomial ax + mx + nx + c by grouping.
That means, factor a common factor out of the first two terms, and another common factor out of the last 2 terms.
5. Factor out the common factor.

Answer 10

Let's use your first problem as an example.

Answer 11

ok...

Answer 12

\(4x^2 -4x+1=36\)

For now we just want to factor the left side.

Answer 13

yea m having trouble with that

Answer 14

wait a second can i dived by 4 to get rid of the coefficient before x^2???

Answer 15

That's why we're going to do it together.

Compared to
\(\color{red}{a}x^2 + \color{green}{b}x + \color{blue}{c}\), you see that for the trinomial
\(\color{red}{4}x^2 \color{green}{- 4}x + \color{blue}{1}\),
\(\color{red}{a = 4}\), \(\color{green}{b = -4}\) and \(\color{blue}{c = 1}\).

(I hope the colors make it clear.)

Answer 16

No, don't divide by 4.

Answer 17

Since we have a = 4, and c = 1, we multiply ac to get ac = 4(1) = 4.

Answer 18

Now we need two factors of 4 that add to b, -4.

Answer 19

Those would be -2 and -2, since (-2)(-2) = 4, and (-2) + (-2) = -4.

Answer 20

Now we break up the middle term into those two numbers we found:
\(4x^2 - 4x + 1\)

\( = 4x^2 - 2x - 2x + 1 \)

Now we factor by grouping.

Answer 21

so the answer would be x =8 or 4

Answer 22

I have no idea about the answer. We are still factoring.

\(4x^2 - 2x - 2x + 1 \)

\(=2x(2x - 1) - 1(2x - 1) \)

Now we see that the line above has a common factor. See below in color:

\(=\color{blue}{2x}\color{red}{(2x - 1)} \color{green}{- 1}\color{red}{(2x - 1)} \)

Now we factor out the red common factor to find the factorization:

\( \color{red}{(2x - 1)}(\color{blue}{2x} \color{green}{- 1}) \)

The final factorization is:

\( (2x - 1)^2 \)

Answer 23

Now that the left side is factored, that means we have
\( (2x - 1)^2 = 36\)

In a previous post of yours, I wrote that when you have an equation of the form
\(x^2 = b\), you solve it by doing the following:
\(X = \pm \sqrt{b} \)

Now we apply this here:

\( 2x - 1 = \pm \sqrt{36} \)

\(2x - 1 = 6\) or \(2x - 1 = -6 \)

\(2x = 7\) or \(2x = -5\)

\(x = \dfrac{7}{2} \) or \(x = -\dfrac{5}{2} \)

Answer 24

ok i get it thanks for the next one how would u factor it?

Answer 25

\(x^2 -\dfrac{2}{3}x+\dfrac{1}{9}=1 \)

My suggestion for the second one is to first multiply both sides by 9 to get rid of all denominators.

Answer 26

@ganeshie8

Answer 27

\(9(x^2 -\dfrac{2}{3}x+\dfrac{1}{9})=9(1)\)

\(9x^2 -6x+1=9\)

\((3x - 1)^2 = 9 \)

\( 3x - 1 = \pm\sqrt{9} \)

\(3x - 1 = 3\) or \(3x - 1 = -3\)

3x = 4 or 3x = -2

\(x = \dfrac{4}{3}\) or \(x = -\dfrac{2}{3} \)