SSC CGL Solution Set 22 Algebra 7 Part 2 | Difficult algebra | Rich patterns | Quick solutions

Welcome friends. This is Atanu Chaudhuri of Suresolv. Continuing from the first part of the video in this second … part we’ll solve the question 6 to question 10 of SSC CGL … Algebra Solution set 7 from suresolv.com. Let’s solve the first problem. Question 6 is, if x(x-3)=-1, then the value of x^3(x^3-18) will be, option a:2, option b:0, option c:1 and option d:-1. Comparing the target expression with the given expression, we decide to raise the given expression to its cube straightaway. To get, x^3(x-3)^3=-1 We will expand (x-3)^3 in a compact form. So this results in, x^3[x^3-9x(x-3)-27]=-1. or, x^3[x^3+9 -27]=-1 or, x^3(x^3-18)=-1 We have got the value of the target expression. Answer option:d. The problem is not that difficult as it looked like. Let’s solve the next problem. Question 7 is, if x=√5+2, then the value of (2x^2-3x-2)/(3x^2-4x-3) is, option a:0.525, option b:0.625, option c:0.785 and option d:0.185. The given value is a surd variable and the choice values … do not have surd at all. So we have to first eliminate the surd. In this case it will be easy to do because the difference of … the squares of the two terms √5 and 2 is just 1. In this situation we can eliminate the the surd by forming x-1/x. this is equal to (√5+2)-1/(√5+2} Rationalizing the denominator we have=√5+2-(√5-2)=4 Our surd is eliminated. Next target is now to form the expression x-1/x in both numerator and denominator of the target expression. That is also easy to do. We divide both the numerator and denominator by x and … take 1st and 3rd term together for both numerator and denominator. For the numerator we get 2(x-1/x)-3, we divided by x. And for the denominator we get 3(x-1/x)-4 this is equal to (2×4-3) divided by (3×4-4)=5/8 which is equal to 0.625. Answer is option:b 0.625 We have used two techniques here. First we have eliminated the surd by forming x-1/x and … second we have formed the expression x-1/x in both the … numerator and the denominator by combining the … first term and the third term in both the numerator and the denominator. These two terms have a special pattern. For both 2x^2 and -2 the coefficients are same which is 2 and the difference in powers of x is also 2. The same rule is true for 3x^2 and -3. That’s why we could form the expression x-1/x out of these … pairs of terms. Let’s solve the next problem. Question 8 is, if p/a+q/b+r/c=1, and a/p+b/q+c/r=0, where p, q, r, a, b and c are non-zero the value of p^2/a^2+q^2/b^2+r^2/c^2 is option a:2, option b:-1, option c:0 and option d:1. The problem statement is quite large. As many as 6 variables are also involved. It seems to be a complex problem. Then we notice a special pattern in all the three expressions. p/a, q/b and r/c appear in all the three expressions standalone as if … these are independent variables. In this situatuation we can very well substitute x for p/a, y for q/b and z for r/c This will simplify all three expressions greatly. The first expression becomes just x+y+z=1. The second expression becomes, 1/x+1/y+1/z=0. Simplifying xy +yz+zx divided by xyz=0 which means, xy+yz+zx=0. And the target expression is transformed to x^2+y^2+z^2. There is no trace of p/a, q/b or r/c. If you solve this problem, you will solve the original problem. That is the advantage of dummy variable substitution. From the well-known formula of (x+y+z)^2 we get (x+y+z)^2=x^2+y^2+z^2 2(xy+yz+zx) This is zero. And this is 1 So x^2+y^2+z^2 which is p^2/a^2+q^2/b^2+r^2/c^2=1 Answer is option d:1 At the end the problem is not so difficult as it seemed. Let’s solve the next problem. Question 9 is, if equation 2x^2-7x+12=0 has … two roots α and β, then the value of α/β+β/α is option a:1/24, option b:7/24, option c:7/2 and option d:97/24. You have to how to know the values of α+β and αβ from a … quadratic equation when the two roots of the quadratic … equation are α and β. But to simplify matters we divide the given expression by 2 getting, x^2-7x/2+6=0 So that we can express the equation straightaway in … terms of its roots as (x-α)(x-β)=0 From this you can easily see that α+β=-7/2 the middle term and αβ=6, the numeric term. Now we will simplify the target expression. Combining the two terms of the target expression we get, (α^2+β^2)/αβ We will add 2 to this expression to form (α+β)^2 in … the numerator plus and compensate by -2=(α+β)^2 in the numerator divided by αβ -2=49/4 in the numerator divided by 6 -2=49/24-2=1/24 Answer is option:a:1/24 Not a difficult problem Let’s solve the last question Question 10 is, If a-b=3, and a^3-b^3=117, then absolute value of a+b is option a:5, option b:7, option c:3 and option d:9. We will straightaway expand a^3-b^3 a^3-b^3=(a-b)(a^2+ab+b^2) This is a standard two-factor expansion of a^3-b^3 or, 117=3[(a-b)^2+3ab]=3(9+3ab). From this we will get the value of ab 117/3 is 39 39-9=3ab It follows ab=10. With values of a-b and ab we can get the absolute value of a+b easily. (a+b)^2=(a-b)^2+4ab=9+40=49 So a+b=±7 absolute value will be seven Answer option:b:7 Overall none of the 5 problems is easy but we could solve them quickly by pattern discovery … and using suitable methods. If you want to read the detailed solutions, refer to the article in suresolv.com. We have given the link in the description of this video. Thanks for watching.

Bernard Jenkins

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  1. I have added voice transcriptions to the video. To watch the video with the transcriptions click on the CC button on the bottom right of the screen.

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