How do you solve ${(2 s - 1)}^{2} = 225$ $(2s-1)^{2}=225$ ?

Question

How do you solve ${(2 s - 1)}^{2} = 225$ $(2s-1)^{2}=225$ ?

2 Answers

Impact of this question

5440 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-18T06:07:32

$x = - 7$ $x=-7$ or $x = 8$ $x=8$

Explanation:

There is a rule that states ${(a + b)}^{2} = a^{2} + 2 a b + b^{2}$ $(a+b)^2=a^2+2ab+b^2$

In this case, $a$ $a$ is $2 s$ $2s$ and $b$ $b$ is $- 1$ $-1$ .

Therefore

$4 s^{2} - 4 s + 1 = 225$ $4s^2-4s+1=225$

Subtract $225$ $225$ from both sides.

$4 s^{2} - 4 s - 224 = 0$ $4s^2-4s-224=0$

Divide by $4$ $4$ .

$s^{2} - s - 56 = 0$ $s^2-s-56=0$

You can factorise this as follows :

$(x + 7) (x - 8)$ $(x+7)(x-8)$

Set each factor equal to $0$ $0$ .

$x + 7 = 0$ $x+7=0$
$x = - 7$ $x=-7$

$x - 8 = 0$ $x-8=0$
$x = 8$ $x=8$

Answer 2 · 2017-06-18T07:26:30

$s = 8 or s = - 7$ $s = 8 or s = -7$

Explanation:

Although this equation leads to a quadratic trinomial, it is in the form $x^{2} = c$ $x^2 =c" "$ so we can just find the square root of both sides,

${(2 s - 1)}^{2} = 225$ $(2s-1)^2 = 225$

$2 s - 1 = \pm \sqrt{225}$ $2s-1 = +-sqrt225$

$2 s = \pm \sqrt{225} + 1$ $2s = +-sqrt225+1$

$s = \frac{+ 15 + 1}{2} = 8$ $s = (+15+1)/2 = 8$

$s = \frac{- 15 + 1}{2} = - 7$ $s = (-15+1)/2 = -7$

Science

Math

Humanities

... and beyond

How do you solve ${(2 s - 1)}^{2} = 225$ $(2s-1)^{2}=225$ ?

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you solve (2s−1)2=225(2s-1)^{2}=225?

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question

How do you solve ${(2 s - 1)}^{2} = 225$ $(2s-1)^{2}=225$ ?