SSC CGL Solution Set 58 Algebra 14 Part 1 | Smart methods | Difficult algebra | Quick solutions

SSC CGL Solution Set 58 Algebra 14 Part 1 | Smart methods | Difficult algebra | Quick solutions


Welcome friends. This is Atanu Chaudhuri of Suresolv. In this first part, we will now solve the first 5 … questions of SSC CGL Solution set 58 on Algebra 14 … from suresolv.com. A few of the problems may seem to be difficult, but we’ll see how to solve them quickly in a few steps. Let’s solve the first problem. Question 1 is, If x=5−√21, then the value of √x/[√(32−2x)- √21] is, option a: (1/√2)( √3-√7) , option b: (1/√2)( √7+√3) option c: (1/√2)( √7-√3), and option d: (1/√2)( 7-√3) √x=√(5−√21) is a double square root surd as √21 … comes under a second square root. To solve the problem, we have to express x as a … whole square expression to free √x of its outer square … root. This is called double square … root surd simplification. Let us do it first. If you consider a general expansion of a perfect square, a^2+2ab+b^2, you would notice that the … middle term invariably comes with the coefficient of 2. In our case (5−√21) is the final squared result, but the middle term of surd is without its coefficient of 2. In this case we use the technique of forcibly … introducing the coefficient 2 by multiplying and dividing the … expression by 2 getting, (5−√21)=½(10-2√21) Now √21 is to be split into two factors in such a way that the … sum of their squares is 10. It is easy to see that √7 and … √3 are these two factors of √21. So we will express √x in a new way now as, √x=√[1/2((√7)^2-2. √7. √3+(√3)^2)]=√[1/2(√7-√3)^2]=1/√2(√7-√3). We have freed up the value of √x of its outer square root. Now the value is a normal surd expression. But we won’t substitute this value of √x in the target … expression straightaway. We’ll simplify the target expression as far as possible first. To simplify the target expression we would take up … the denominator first and substitute the given value of … √x. We would get the … denominator, say, D as, D=√(32−2x) – √21=√[32−2(5−√21)]- … √21=√[22+2√21]-√21 The first term is again a double square root surd, but this time the surd middle term has its coefficient of 2. It is easy to identify 22 as the sum of squares of √21 and 1 … and so we express the simplified denominator as, D=√(√21+1)^2-√21=1. The value of the target expression is then, E=√x=1/√2(√7-√3). Answer is: Option c: 1/√2(√7-√3). It is not difficult to solve this problem in mind if you know … the concepts and methods. Let’s solve the next problem. Question 2 is. If a+b+c+d=4, then find the value of 1/(1−a)(1−b)(1−c)+1/(1−b)(1−c)(1−d)+1/(1−c)(1−d)(1−a)+1/(1−d)(1−a)(1−b) … is, option a:1, option b:4, option c:5, and option d:0 The main question is — how to use the given expression to … get a useful relation between the four factors (1-a), (1-b), (1-c) and (1-d) which make up … the target expression. It takes a just moment to identify easily that the key lies … in the numeric term 4 on the RHS of the given expression. Just share 4 equally between the four variables and you will get, (1-a)+(1-b)+(1-c)+(1-d)=0. This is the key pattern discovery for solving the … problem immediately. Just add the four algebraic fraction terms like a normal … fraction addition, and you will have the sum of … the four factors in the numerator with their product in … the denominator. E=[(1-a)+(1-b)+(1-c)+(1-d)]/ (1-a)(1-b)(1-c)(1-d)=0 As the sum of the factors in the numerator is 0. Solved easily in mind. Answer is: Option d : 0. Let’s solve the 3rd problem. Question 3 is. If a(2+√3)=b(2−√3)=1, then the positive value of 1/(a^2+1)+1/(b^2+1) is, option a:1, option b: 4, option c:9, and option d:−5 We identify that the two surd expressions are suitable for … inversion and rationalization as, 2+√3=1/(2−√3), and vice versa. We have multiplied and divided 2+√3 by 2-√3. Interestingly, you will get this result for any … two-term surd expression for which the difference of … squares of the two terms is 1. In this case also, 2^2-(√3)^2=1. With this result it follows, 2+√3+1/(2+√3)=… 2-√3+1/(2-√3)=( 2+√3)+( 2-√3)=4 This special result provides the quick solution when we … take out a and b as a factor from the first and second … denominator of the target expression, E=1/a[1/(a+1/a)]+ 1/b[1/(b+1/b)] From the given expression we have already, a=1/(2+√3)=2-√3, and b=1/(2−√3)=2+√3, So, a+1/a=b+1/b=4. Substituting this result in the target expression, E=1/4(1/a+1/b)=1/4(2+√3+2-√3)=1. Answer is: Option a: 1. Easily solved in mind. Let’s solve the next problem. Question 4 is. If … x/(xa+yb+zc)=y/(ya+zb+xc)=z/(za+xb+yc), and x+y+z≠0 then each ratio … can be expressed as, option a:1/(a+b−c), option b: 1/(a+b+c), option c: 1/(a−b−c), and option d: 1/(a−b+c). The given expression forms a chain of three expressions … and so it needs application of Chained equation treatment technique. By this technique, an additional dummy constant … is appended to the end of the chain giving, x/(xa+yb+zc)=y/(ya+zb+xc)=z/(za+xb+yc)=p. The advantage is—now we are able to form three … independent equations with the dummy constant as, x/(xa+yb+zc)=p, Or, x=p(xa+yb+zc). Similarly, y/(ya+zb+xc)=p, Or, y=p(ya+zb+xc), and z/(za+xb+yc)=p, Or, z=p(za+xb+yc). The solution is just a step away now. Add up the three equations and you will get, x+y+z=p(a+b+c)(x+y+z), Or, p=1/(a+b+c), as (x+y+z)≠0. Answer is: Option b: 1/(a+b+c). Again solved quickly in only a few steps. Last problem now. Question 5 is. If 3(a^2+b^2+c^2)=(a+b+c)^2, then the relation between a, b and c is, option a:a=b=c, option b: a≠b=c, option c: a=b≠c, and option d: a≠b≠c Expanding the RHS we will get one number of … (a^2+b^2+c^2) reducing three such factors in the LHS to 2. Let’s show you the transformation by expanding (a+b+c)^2, 3(a^2+b^2+c^2)=a^2+b^2+c^2+2(ab+bc+ca), Or, 2(a^2+b^2+c^2)=2(ab+bc+ca). Now it is a straightforward rearrangement of terms and … application of the important algebraic concept of zero sum … of square expressions. Collecting the like terms together we have, (a^2-2ab+b^2)+(b^2-2bc+c^2)+(c^2-2ca+a^2)=0 Or, (a−b)^2+(b−c)^2+(c−a)^2=0. With real a, b and c, for their sum to be zero each of the square expressions … must be zero. So, (a−b)=(b−c)=(c−a)=0, Or, a=b=c. Answer is: Option a: a=b=c. You can solve this problem also easily in mind if you … identify the concept and know how to apply the method. To read the detailed solutions refer to the corresponding … article in our website suresolv.com. The link is in the description of this video. And for latest updates, subscribe to our channel. Thanks for watching. [No Text] [No Text]

One thought on “SSC CGL Solution Set 58 Algebra 14 Part 1 | Smart methods | Difficult algebra | Quick solutions

  • August 14, 2019 at 9:35 am
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    Added transctiptions

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