hence thus 53-12√/10=45+8-2√45x8 =(√45-8)2; ... √53-12√10=√45−√/8 =3√5–2√2. Example 3. Find the square root of a+b+ √2ab+b2. Rewrite the binomial so that the surd part has a coefficient 2; We have now to find two quantities whose sum is (a+b) and NOTE. The student should notice that when the coefficient of the surd part of the binomial is unity he can make this coefficient 2 if he will also multiply the quantity under the radical by 4. 283. Assuming Va+√b=x+√y, the method of Art. 278 gives us Va√b=x− √y• EXAMPLES XXXII. f. Find the square root of each of the following binomial surds: 20. 1+a2+(1+a2+a1)§. 16. 2a+2√a2 — b2. 17. ax-2a√ax-a2. 18. a+x+√2ax+x2. 19. 2a-√3a2 — 2ab—b2. Find the fourth roots of the following binomial surds: 284. Sometimes equations are proposed in which the unknown quantity appears under the radical sign. Such equations are very varied in character and often require special artifices for their solution. Here we shall only consider a few of the simpler cases, which can generally be solved by the following method: Bring to one side of the equation a single radical term by itself: on squaring both sides this radical will disappear. By repeating this process any remaining radicals can in turn be removed. Example 1. Solve 2x-4x-11=1. Transposing, 2x-1=4x-11. Square both sides; then 4x-4√√x+1=4x − 11, Here we must cube both sides; thus x-5=1331; whence x=1336. Example 3. Solve √x+5+ √3x+4= √12x+1. x+5+3x+4+2 √√(x+5) (3x+4)=12x+1. Transposing and dividing by 2, If we proceed to verify the solution by substituting these values in the original equation, it will be found that it is satisfied by x=4, but 1 13 not by x= But this latter value will be found on trial to satisfy the given equation if we alter the sign of the second radical; thus √x+5−√3x+4= √12x+1. On squaring this and reducing, we obtain · √(x+5) (3x+4)=4x − 4 ..... .(2); and a comparison of (1) and (2) shews that in the next stage of the work the same quadratic equation is obtained in each case, the roots of which are 4 and 1 as already found. 13 From this it appears that when the solution of an equation requires that both sides should be squared, we cannot be certain without trial which of the values found for the unknown quantity will satisfy the original equation., In order that all the values found by the solution of the equation may be applicable it will be necessary to take into account both signs of the radicals in the given equation. 10. √4x-7x+1=2x-1. 14. 10-√25+9x=3√x. 16. 18. 11. √x+25=1+√x. 9x-8=3√x+4-2, 285. When radicals appear in a fractional form in an equation, we must clear of fractions in the ordinary way, combining the irrational factors by the rules already explained in this chapter. Multiplying across, 6x+25x - 66=6x+3/x, 25/x-3x=66, 22/x=66, √x=3, x=9. |