Question

Help please 50 points!​

Answer 1

A and B

Explanation:

we would like to solve the following equation:

`\rm\displaystyle 5 - 2x = \sqrt{ {2x}^(2) + x - 1 } - x`

to do so isolate -x to the left hand side and change its sign:

`\rm\displaystyle 5 - 2x + x = \sqrt{ {2x}^(2) + x - 1 }`

simplify addition:

`\rm\displaystyle 5 - x = \sqrt{ {2x}^(2) + x - 1 }`

square both sides:

`\rm\displaystyle (5 - x {)}^(2) = (\sqrt{ {2x}^(2) + x - 1 } {)}^(2)`

simplify square of the right hand side:

`\rm\displaystyle (5 - x {)}^(2) = {2x}^(2) + x - 1`

use (a-b)²=a²-2ab+b² to expand the left hand side:

`\rm\displaystyle {x}^(2) - 10x + 25= {2x}^(2) + x - 1`

swap the equation:

`\rm\displaystyle {2x}^(2) + x - 1 = {x}^(2) - 10x + 25`

isolate the right hand side expression to the left hand side and change every sign:

`\rm\displaystyle {2x}^(2) + x - 1 - {x}^(2) + 10x - 25 = 0`

simplify:

`\rm\displaystyle {x}^(2) + 11x - 26= 0`

rewrite the middle term as 13x-2x:

`\rm\displaystyle {x}^(2) + 13x - 2x - 26= 0`

factor out x:

`\rm\displaystyle x({x}^{} + 13)- 2x - 26= 0`

factor out -2:

`\rm\displaystyle x({x}^{} + 13)- 2(x + 13)= 0`

group:

`\rm\displaystyle (x- 2)(x + 13)= 0`

by Zero product property we obtain:

`\displaystyle \begin{cases}x- 2 = 0\\ x + 13= 0 \end{cases}`

solve for x:

`\displaystyle \begin{cases}x = 2\\ x = - 13 \end{cases}`

to check for extraneous solutions we can define the domain of equation recall that a square root of a function is always greater than or equal to 0 therefore

`\rm\displaystyle 5 - x \geq0`

solve the inequality for x:

`\rm\displaystyle x \leqslant 5`

since 2 and -13 is less than 5 both solutions are valid for x hence,

`\displaystyle \begin{cases}x _(1) = 2\\ x_(2) = - 13 \end{cases}`

and we're done!

Answer 2

x=-13

x=2

Answer:

Solution given:

5-2x=
`√(2x²+x-1)-x`

keep square root alone

5-2x+x=
`√(2x²+x-1)`

5-x=
`√(2x²+x-1)`

squaring both side

(5-x)²=2x²+x-1

25-10x+x²=2x²+x-1

2x²+x-1-25+10x-x²=0

x²+11x-26=0

doing middle term factorisation

for this

coefficient of first and last terms should be multiplied

here

1*26=26

factor product

26=2*13*1

remember that check the sigh of last term

in here it is - so

we need to get 11 by subtracting factor of 26

we get

13-2=11

keep 13-2 value in place of 11

x²+(13-2)x-26

now

distribute

x²+13x-2x-26=0

take common from each two term

x(x+13)-2(x+13)=0

(x+13)(x-2)=0

either

x=-13

or.x=2