Question

Help please 50 points!​

Answer 1

x=-13

x=2

Answer:

Solution given:

5-2x=$\sqrt{2x²+x-1}-x$

keep square root alone

5-2x+x=$\sqrt{2x²+x-1}$

5-x=$\sqrt{2x²+x-1}$

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

Answer 2

Answer:

A and B

Step-by-step 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!