The first thing you want to do in radical equations is get the radical on one side of the equation. Today is our lucky day, because that has already been done for us.
Next step is to square both sides to get rid of the radical:
sqrt(2x+7)=x+3√2x+7=x+3
(sqrt(2x+7))^2=(x+3)^2(√2x+7)2=(x+3)2
->2x+7=x^2+6x+9→2x+7=x2+6x+9
Now we have to combine like terms and set the equation equal to 00:
2x+7=x^2+6x+92x+7=x2+6x+9
0=x^2+(6x-2x)+(9-7)0=x2+(6x−2x)+(9−7)
->0=x^2+4x+2→0=x2+4x+2
Unfortunately, this quadratic equation does not factor, so we'll have to use the quadratic formula:
x=(-b+-sqrt(b^2-4ac))/(2a)x=−b±√b2−4ac2a
With a=1a=1, b=4b=4, and c=2c=2, our solutions are:
x=(-(4)+-sqrt((4)^2-4(1)(2)))/(2(1))x=−(4)±√(4)2−4(1)(2)2(1)
x=(-4+-sqrt(16-8))/2x=−4±√16−82
x=-4/2+-sqrt(8)/2x=−42±√82
->x=-2+-sqrt(2)→x=−2±√2
(Note that sqrt(8)/2=(2sqrt(2))/2=sqrt2√82=2√22=√2)
We have our solutions: x=-2+sqrt2~~-0.586x=−2+√2≈−0.586 and x=-2-sqrt2~~-3.414x=−2−√2≈−3.414. But because this is an equation involving radicals, we need to double-check our solutions.
Solution 1: x~~-0.586x≈−0.586
sqrt(2x+7)=x+3√2x+7=x+3
sqrt(2(-0.586)+7)=-0.586+3√2(−0.586)+7=−0.586+3
2.414=2.414->2.414=2.414→ Solution checks
Solution 2: x~~-3.414x≈−3.414
sqrt(2x+7)=x+3√2x+7=x+3
sqrt(2(-3.414)+7)=-3.414+3√2(−3.414)+7=−3.414+3
.415!=-.414->.415≠−.414→ Extraneous solution
As you can see, only one of our solutions work: x=-2+sqrt2x=−2+√2.
