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# What is the solution of the equation?

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What is the solution of the equation?

May 26, 2018

#1
+988
+2

$$\sqrt{2x-5}+4=x\\ \sqrt{2x-5}+4-4=x-4\\ \sqrt{2x-5}=x-4\\ (\sqrt{2x-5})^2=(x-4)^2\\ 2x-5=x^2-8x+16\\ x^2-10x+21=0\\ (x-7)(x-3)=0\\ x_1=7, x_2 = 3$$

We plug the solutions to verify:

$$\sqrt{2\cdot7-5}+4=7\\ \sqrt9+4=7\\ 3+4=7\\$$

Seven works!

$$\sqrt{2\cdot3-5}+4=3\\ \sqrt1+4=3\\ 5\ne3\\$$

Three doesn’t work.

Therefore, the answer you seek is 7!

I hope this helped,

Gavin

May 26, 2018

#1
+988
+2

$$\sqrt{2x-5}+4=x\\ \sqrt{2x-5}+4-4=x-4\\ \sqrt{2x-5}=x-4\\ (\sqrt{2x-5})^2=(x-4)^2\\ 2x-5=x^2-8x+16\\ x^2-10x+21=0\\ (x-7)(x-3)=0\\ x_1=7, x_2 = 3$$

We plug the solutions to verify:

$$\sqrt{2\cdot7-5}+4=7\\ \sqrt9+4=7\\ 3+4=7\\$$

Seven works!

$$\sqrt{2\cdot3-5}+4=3\\ \sqrt1+4=3\\ 5\ne3\\$$

Three doesn’t work.

Therefore, the answer you seek is 7!

I hope this helped,

Gavin

GYanggg May 26, 2018