Solve for x by factoring or by using the quadratic equation
0=(x-3)(x-7)
Therefore, the two possible answers are x=3 and x=7
However, equations that involve variables underneith square roots tend to look like one half of a quadratic curve that has been rotated 90 degrees to one side.
There can only be one answer and the other would be an extraneous root. Verify x=3 by substituting it back into the original formula:
$\sqrt{2\left(3\right)-5}-4=-3$√2(3)−5−4=−3
$\sqrt{1}-4=-3$√1−4=−3
-3=-3
Therefore, x=3 is the one correct solutions. If you were to substitute x=7 into this equation, the left hand side would not equal the right hand side and concluded that it is an extraneous root and not the solution to the problem.
Attached Graph:
6 years ago
Answered By Sujalakshmy V
$\sqrt{2x-5}-4=-x$√2x−5−4=−x
First step in solving a radical equation is to find the non-permissible values of x
This can be found out as 2x-5 $\ge$≥ 0
which gives 2x $\ge$≥ 5
or x $\ge$≥ 5/2
Now to solve the equation, isolate the radical part and rearrange the equation as
$\sqrt{2x-5}=-x+4$√2x−5=−x+4
Next step is to remove the radical signs, so square both sides of the equation which gives
$\left(\sqrt{2x-5}\right)^2$(√2x−5)2 =(-x+4)2
2x-5 = x2 -8x +16
Arrange the equation to form a quadriatic equation
x2-10x +21 =0
On factoring the above equation,
(x-3)(x-7)=0
which gives x =3 , and x= 7
Both the values of x satisfies the non permissible conditions .
Verification :
when x =3
Left side : Right side:
$\sqrt{2x-5}$√2x−5 - 4 -x
$\sqrt{2\times3-5}$√2×3−5 - 4 -3
$\sqrt{6-5}$√6−5 - 4 -3
1 - 4 -3
-3 -3
LS = RS Therefore x=3 is a solution
When x=7
Left side Right side
$\sqrt{2x-5}$√2x−5 - 4 -x
$\sqrt{2\times7-5}$√2×7−5 - 4 -7
$\sqrt{14-5}$√14−5 - 4 -7
$\sqrt{9}$√9 - 4 -7
3 - 4 -7
-1 $\ne$≠ -7
LS $\ne$≠ RS
Therefore x=7 cannot be a solution or it is an extraneous solution.
The solution for the equation $\sqrt{2x-5}-4=-x$√2x−5−4=−x is x=3
6 years ago
Answered By Alireza R
Re-arrange the equation:
$\sqrt{2\times x-5}=-x+4$√2×x−5=−x+4
Square both sides:
$2x-5=x^2-8x+16$2x−5=x2−8x+16
Move everything to one side:
$x^2-10x+21=0$x2−10x+21=0
Solve the quadratic equation:
a=1
b=-10
c=21
x= $\frac{-\left(-10\right)\pm\sqrt{\left(-10\right)^2-4\times\left(1\right)\times\left(16\right)}}{\left(2\times\left(1\right)\right)}$−(−10)±√(−10)2−4×(1)×(16)(2×(1)) =
X= 3 and 7
Please note that 2x-5must be equal or greater than zero, which means $x\ge2.5$x≥2.5
6 years ago
Answered By Corey A
An exercise in algebra and soving for a single variable x:
$\sqrt{2x-5}-4=-x$√2x−5−4=−x
Add 4 to both sides
$\sqrt{2x-5}=-x+4$√2x−5=−x+4
Square both sides to remove square root sign and foil right hand side of equation.
$\left(\sqrt{2x-5}\right)^2=\left(-x+4\right)^2$(√2x−5)2=(−x+4)2
$2x-5=x^2-8x$2x−5=x2−8x +16
Collect like terms and set equal to 0
$0=x^2-10x+21$0=x2−10x+21
Solve for x by factoring or by using the quadratic equation
0=(x-3)(x-7)
Therefore, the two possible answers are x=3 and x=7
However, equations that involve variables underneith square roots tend to look like one half of a quadratic curve that has been rotated 90 degrees to one side.
There can only be one answer and the other would be an extraneous root. Verify x=3 by substituting it back into the original formula:
$\sqrt{2\left(3\right)-5}-4=-3$√2(3)−5−4=−3
$\sqrt{1}-4=-3$√1−4=−3
-3=-3
Therefore, x=3 is the one correct solutions. If you were to substitute x=7 into this equation, the left hand side would not equal the right hand side and concluded that it is an extraneous root and not the solution to the problem.
Attached Graph:
6 years ago
Answered By Sujalakshmy V
$\sqrt{2x-5}-4=-x$√2x−5−4=−x
First step in solving a radical equation is to find the non-permissible values of x
This can be found out as 2x-5 $\ge$≥ 0
which gives 2x $\ge$≥ 5
or x $\ge$≥ 5/2
Now to solve the equation, isolate the radical part and rearrange the equation as
$\sqrt{2x-5}=-x+4$√2x−5=−x+4
Next step is to remove the radical signs, so square both sides of the equation which gives
$\left(\sqrt{2x-5}\right)^2$(√2x−5)2 =(-x+4)2
2x-5 = x2 -8x +16
Arrange the equation to form a quadriatic equation
x2-10x +21 =0
On factoring the above equation,
(x-3)(x-7)=0
which gives x =3 , and x= 7
Both the values of x satisfies the non permissible conditions .
Verification :
when x =3
Left side : Right side:
$\sqrt{2x-5}$√2x−5 - 4 -x
$\sqrt{2\times3-5}$√2×3−5 - 4 -3
$\sqrt{6-5}$√6−5 - 4 -3
1 - 4 -3
-3 -3
LS = RS Therefore x=3 is a solution
When x=7
Left side Right side
$\sqrt{2x-5}$√2x−5 - 4 -x
$\sqrt{2\times7-5}$√2×7−5 - 4 -7
$\sqrt{14-5}$√14−5 - 4 -7
$\sqrt{9}$√9 - 4 -7
3 - 4 -7
-1 $\ne$≠ -7
LS $\ne$≠ RS
Therefore x=7 cannot be a solution or it is an extraneous solution.
The solution for the equation $\sqrt{2x-5}-4=-x$√2x−5−4=−x is x=3