Answer:Together, these imply that the polynomial is non-negative only for an interval of values between its two roots. Applying the quadratic equation tells us that these roots are −0.5−0.5 and 1.51.5. So this polynomial isnon-negative only for the interval [−0.5,1.5][−0.5,1.5].Similarly, we see that the polynomial under the second radical is also quadratic with the coefficient of thequadratic term being negative. It is easy to factor, andwe see that its two roots are 00 and 22. So we know that this second polynomial is only non-negative on theinterval [0,2][0,2].To get a real solution, we know that the arguments of both square roots must be non-negative, so we only need to look for solutions on the interval that is theintersection of [−0.5,1.5][−0.5,1.5] and [0,2][0,2]. So we seek real solutions only in [0,1.5][0,1.5].Now, notice that both of the intervals, [−0.5,1.5][−0.5,1.5] and[0,2][0,2], are centered at x=1x=1. If you know much about parabolas, you may know that their unique maximum (or minimum for parabolas with quadratic term having positive coefficient) occurs at the center of the interval spanned by its roots. So we know that both polynomials are maximized at x=1x=1. As the square root function ismonotonically increasing, it follows that both square roots are also maximized at x=1x=1. And finally, since the square roots are added, it follows that their sum ismaximized at x=1x=1. So by plugging in x=1x=1 to theleft-hand-side, we see that the expression on the left-hand-side of this equality has maximum value of 55which occurs at x=1x=1.What about the right-hand-side? Well, it is a quadraticpolynomial whose leading coefficient is positive which means it has a unique minimum. We can complete thesquare to write the this polynomial as 4x2−8x+9=4(x2−2x+1)+5=4(x−1)2+54x2−8x+9=4(x2−2x+1)+5=4(x−1)2+5. Notice that in this final form, the expression is the sum of two non-negative terms. It is clearly minimized when the 4(x−1)24(x−1)2 term is as small as possible which will be when x=1x=1 making this term zero and the entire right-hand-side equal to 55.So think about what we just learned. The LARGEST possible value of the left-hand-side is 55 and occurs when x=1x=1. And the SMALLEST possible value of theright-hand-side is 55 and occurs when x=1x=1. And just like that, we can conclude not only that x=1x=1 is A SOLUTION (since both sides of the equation are equal for this value), but also that x=1x=1 is the ONLY (real) SOLUTION since for all other (real) values of xx, theleft-hand-side must be smaller than 55, and the right-hand-side must be larger than 55; hence, they cannot be equal.
Explanation: