Learn about Approximate Solutions to Quadratic Equations Approximating solutions. So in this problem, we need to solve and then round to the nearest hundredth. So lets take a look at our first problem, x2=10, so here we’re going to take the square root of both sides and use the positive and negative to show both square roots. So here, we’re going to take x=v10 and then v-10. So now, the best estimate of a v10 would be 3.162. So if I round 3.162 to the nearest hundredth, I get approximately 3.16. So this is if I round to the nearest hundredth. So the exact solution to this problem is v10 and -v10 but the approximate solution would be 3.16 or -3.16. All right, let’s go ahead and move on to our second problem. So here we have 0=-2x2+80, so again, we need to get x by itself. So I'm going to subtract 80 from both sides, so I'm going to start off by subtracting 80 from both sides. So I get -80=-2x2 then I'm going to divide both sides by -2. And then, when I divide -80 by -2, I get a positive 40=x2. So now, I'm going to take the square root of both sides and use the positive or negative to show both square roots. So x is going to equal the v40 or -v40. Now the best estimate of the v40 would be 6.324 and then if I need the directions remember it told me that I need to round to the nearest hundredth. So if I take my 6.324 and round that to the nearest hundredth, I'm going to get about 6.32. So our exact solution for this problem would be v40 and -v40. The approximate solutions would be 6.32 and -6.32. All right, things to keep in mind, remember that if x2 is equal to A and A is a positive real number, then x is the vA and -vA. We can use the square roots of perfect squares to help estimate the square roots of other numbers. And then, if A is not a perfect square is between two perfect squares that are close to A, so lets give it an example for that because that kind of sound a little confusing. So what we’re saying is that 40 between the perfect squares 36 and 49, so 40 is not a perfect square. But, we could say it’s close to 36 or 49, so we can use these perfect squares to help us find an approximate.