Identities

Q1: Multiple Choice Type:

i. If \(x-\frac{1}{x}=3\), the value of \(x^2+\frac{1}{x^2}\) is:

Step 1: Recall the identity: \((x-\frac{1}{x})^2 = x^2 – 2 + \frac{1}{x^2}\)
Step 2: Substitute: \(3^2 = x^2 – 2 + \frac{1}{x^2}\)
Step 3: \(9 = x^2 + \frac{1}{x^2} – 2\)
Step 4: \(x^2 + \frac{1}{x^2} = 9 + 2 = 11\)
Answer: c. 11

ii. If \(a+b=7\) and \(ab=10\); the value of \(a^2+b^2\) is equal to

Step 1: Use the identity \( (a+b)^2 = a^2 + 2ab + b^2 \)
Step 2: \(7^2 = a^2 + b^2 + 2\cdot10\)
Step 3: \(49 = a^2 + b^2 + 20\)
Step 4: \(a^2 + b^2 = 49 – 20 = 29\)
Answer: a. 29

iii. The value of \(\frac{95\times95-5\times5}{95-5}\) is equal to:

Step 1: Recognize the identity \(p^2 – q^2 = (p-q)(p+q)\)
Step 2: \(95^2 – 5^2 = (95-5)(95+5) = 90 \cdot 100\)
Step 3: Divide by \(95-5 = 90\): \(\frac{90\cdot100}{90} = 100\)
Answer: a. 100

iv. If \(a-b=1\) and \(a+b=3\), the value of ab is:

Step 1: Use the identity \((a+b)^2 – (a-b)^2 = 4ab\)
Step 2: \((3)^2 – (1)^2 = 4ab\)
Step 3: \(9 – 1 = 4ab \Rightarrow 8 = 4ab\)
Step 4: \(ab = \frac{8}{4} = 2\)
Answer: b. 2

v. If \(x+\frac{1}{x}=2\), the value of \(\left(x^3+\frac{1}{x^3}\right)-\left(x^2+\frac{1}{x^2}\right)\)is:

Step 1: Use identity \( x^3 + \frac{1}{x^3} = (x+\frac{1}{x})^3 – 3(x+\frac{1}{x}) \)
Step 2: \( x^3 + \frac{1}{x^3} = 2^3 – 3\cdot 2 = 8 – 6 = 2\)
Step 3: Use identity \( x^2 + \frac{1}{x^2} = (x+\frac{1}{x})^2 – 2 = 2^2 – 2 = 4 – 2 = 2\)
Step 4: Compute \(\left(x^3+\frac{1}{x^3}\right) – \left(x^2+\frac{1}{x^2}\right) = 2 – 2 = 0\)
Answer: a. 0

Q2: If \(a+b=5\) and \(ab=6\), find \(a^2+b^2\).

Step 1: Recall the identity: \((a+b)^2 = a^2 + 2ab + b^2\)
Step 2: Rearrange to find \(a^2+b^2\):
\(a^2+b^2 = (a+b)^2 – 2ab\)
Step 3: Substitute the given values: \(a^2+b^2 = 5^2 – 2\cdot6\)
Step 4: Simplify: \(a^2+b^2 = 25 – 12 = 13\)
Answer:13

Q3: If \(a-b=6\) and \(ab=16\), find \(a^2+b^2\)

Step 1: Recall the identity: \((a-b)^2 = a^2 – 2ab + b^2\)
Step 2: Rearrange to find \(a^2+b^2\):
\(a^2 + b^2 = (a-b)^2 + 2ab\)
Step 3: Substitute the given values: \(a^2 + b^2 = 6^2 + 2\cdot16\)
Step 4: Simplify: \(a^2 + b^2 = 36 + 32 = 68\)
Answer:68

Q4: If \(a^2+b^2=29\) and \(ab=10\); find:

i. Find \(a + b\)

Step 1: Recall identity: \((a+b)^2 = a^2 + 2ab + b^2\)
Step 2: Substitute given values: \((a+b)^2 = 29 + 2\cdot10 = 29 + 20 = 49\)
Step 3: Take square root: \(a+b = \sqrt{49} = \pm 7\)
Answer:\(\pm 7\)

ii. Find \(a – b\)

Step 1: Recall identity: \((a-b)^2 = a^2 – 2ab + b^2\)
Step 2: Substitute given values: \((a-b)^2 = 29 – 2\cdot10 = 29 – 20 = 9\)
Step 3: Take square root: \(a-b = \sqrt{9} = \pm 3\)
Answer:\(\pm 3\)

Q5: If \(a^2+b^2=10\) and \(ab=3\); find:

i. Find \(a-b\)

Step 1: Recall identity: \((a-b)^2 = a^2 – 2ab + b^2
Step 2: Substitute given values: \((a-b)^2 = 10 – 2\cdot3 = 10 – 6 = 4\)
Step 3: Take square root: \(a-b = \sqrt{4} = v2\)
Answer:\(\pm 2\)

ii. Find \(a+b\)

Step 1: Recall identity: \((a+b)^2 = a^2 + 2ab + b^2
Step 2: Substitute given values: \((a+b)^2 = 10 + 2\cdot3 = 10 + 6 = 16\)
Step 3: Take square root: \(a+b = \sqrt{16} = \pm 4\)
Answer:\(\pm 4\)

Q6: If \(a+\frac{1}{a}=3\), find: \(a^2+\frac{1}{a^2}\)

Step 1: Recall the identity: \((a + \frac{1}{a})^2 = a^2 + 2 + \frac{1}{a^2}\)
Step 2: Rearrange to find \(a^2 + \frac{1}{a^2}\):
\(a^2 + \frac{1}{a^2} = (a + \frac{1}{a})^2 – 2\)
Step 3: Substitute the given value: \(a^2 + \frac{1}{a^2} = 3^2 – 2 = 9 – 2 = 7\)
Answer:7

Q7: If \(a-\frac{1}{a}=4\), find: \(a^2+\frac{1}{a^2}\)

Step 1: Recall the identity: \((a – \frac{1}{a})^2 = a^2 – 2 + \frac{1}{a^2}\)
Step 2: Rearrange to find \(a^2 + \frac{1}{a^2}\):
\(a^2 + \frac{1}{a^2} = (a – \frac{1}{a})^2 + 2\)
Step 3: Substitute the given value: \(a^2 + \frac{1}{a^2} = 4^2 + 2 = 16 + 2 = 18\)
Answer:18

Q8: If \(a^2+\frac{1}{a^2}=23\), find: \(a+\frac{1}{a}\)

Step 1: Recall the identity: \((a + \frac{1}{a})^2 = a^2 + 2 + \frac{1}{a^2}\)
Step 2: Rearrange to find \(a + \frac{1}{a}\):
\((a + \frac{1}{a})^2 = a^2 + \frac{1}{a^2} + 2\)
Step 3: Substitute the given value: \((a + \frac{1}{a})^2 = 23 + 2 = 25\)
Step 4: Take square root: \(a + \frac{1}{a} = \sqrt{25} = \pm 5\)
Answer:\(\pm 5\)

Q9: If \(a^2+\frac{1}{a^2}=11\), find: \(a-\frac{1}{a}\)

Step 1: Recall the identity: \((a – \frac{1}{a})^2 = a^2 + \frac{1}{a^2} – 2\)
Step 2: Substitute the given value: \((a – \frac{1}{a})^2 = 11 – 2 = 9\)
Step 3: Take square root: \(a – \frac{1}{a} = \sqrt{9} = \pm 3\)
Answer:\(\pm 3\)

Q10: If \(a+b+c=10\) and \(a^2+b^2+c^2=38\), find: \(ab+bc+ca\)

Step 1: Recall the identity: \((a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab+bc+ca)\)
Step 2: Substitute the given values:
\((10)^2 = 38 + 2(ab+bc+ca)\)
Step 3: Simplify: \(100 = 38 + 2(ab+bc+ca)\)
Step 4: Solve for \(ab+bc+ca\):
\(2(ab+bc+ca) = 100 – 38 = 62\)
\(ab+bc+ca = \frac{62}{2} = 31\)
Answer:31

Q11: Find: \(a^2+b^2+c^2\), if \(a+b+c=9\) and \(ab+bc+ca=24\)

Step 1: Recall the identity: \((a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab+bc+ca)\)
Step 2: Rearrange to find \(a^2 + b^2 + c^2\):
\(a^2 + b^2 + c^2 = (a+b+c)^2 – 2(ab+bc+ca)\)
Step 3: Substitute the given values:
\(a^2 + b^2 + c^2 = 9^2 – 2(24) = 81 – 48 = 33\)
Answer:33

Q12: Find: \(a+b+c\), if \(a^2+b^2+c^2=83\) and \(ab+bc+ca=71\)

Step 1: Recall the identity: \((a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab+bc+ca)\)
Step 2: Substitute the given values:
\((a+b+c)^2 = 83 + 2(71) = 83 + 142 = 225\)
Step 3: Take square root: \(a+b+c = \sqrt{225} = \pm 15\)
Answer:\(\pm 15\)

Q13: If \(a+b=6\) and \(ab=8\), find: \(a^3+b^3\)

Step 1: Recall the identity: \(a^3 + b^3 = (a+b)^3 – 3ab(a+b)\)
Step 2: Substitute the given values:
\(a^3 + b^3 = 6^3 – 3(8)(6)\)
Step 3: Simplify:
\(a^3 + b^3 = 216 – 144 = 72\)
Answer:72

Q14: If \(a-b=3\) and \(ab=10\), find \(a^3-b^3\)

Step 1: Recall the identity: \(a^3 – b^3 = (a-b)(a^2 + ab + b^2)\)
Step 2: Find \(a^2 + b^2\) using \((a-b)^2 = a^2 – 2ab + b^2\):
\((a-b)^2 = a^2 – 2ab + b^2\)
\(3^2 = a^2 – 2(10) + b^2\)
\(9 = a^2 + b^2 – 20\)
\(a^2 + b^2 = 29\)
Step 3: Compute \(a^2 + ab + b^2\):
\(a^2 + ab + b^2 = 29 + 10 = 39\)
Step 4: Substitute into \(a^3 – b^3 = (a-b)(a^2 + ab + b^2)\):
\(a^3 – b^3 = 3 \cdot 39 = 117\)
Answer:117

Q15: Find: \(a^3+\frac{1}{a^3}\), if \(a+\frac{1}{a}=5\)

Step 1: Recall the identity: \(a^3 + \frac{1}{a^3} = (a + \frac{1}{a})^3 – 3(a + \frac{1}{a})\)
Step 2: Substitute the given value:
\(a^3 + \frac{1}{a^3} = 5^3 – 3(5)\)
Step 3: Simplify:
\(a^3 + \frac{1}{a^3} = 125 – 15 = 110\)
Answer:110

Q16: Find: \(a^3-\frac{1}{a^3}\), if \(a-\frac{1}{a}=4\)

Step 1: Recall the identity: \(a^3 – \frac{1}{a^3} = (a – \frac{1}{a})^3 + 3(a – \frac{1}{a})\)
Step 2: Substitute the given value:
\(a^3 – \frac{1}{a^3} = 4^3 + 3(4)\)
Step 3: Simplify:
\(a^3 – \frac{1}{a^3} = 64 + 12 = 76\)
Answer:76

Q17: If \(2x-\frac{1}{2x}=4\), find:

i. \(4x^2+\frac{1}{4x^2}\)

Step 1: Recall the identity: \(\left(a – \frac{1}{a}\right)^2 = a^2 + \frac{1}{a^2} – 2\)
Let \(a = 2x\), then:
\((2x – \frac{1}{2x})^2 = (2x)^2 + \frac{1}{(2x)^2} – 2\)
Step 2: Substitute the given value:
\(4^2 = 4x^2 + \frac{1}{4x^2} – 2\)
Step 3: Solve for \(4x^2 + \frac{1}{4x^2}\):
\(16 = 4x^2 + \frac{1}{4x^2} – 2\)
\(4x^2 + \frac{1}{4x^2} = 16 + 2 = 18\)
Answer:18

ii. \(8x^3+\frac{1}{8x^3}\)

Step 1: Recall the identity: \(\left(a – \frac{1}{a}\right)^3 = a^3 – \frac{1}{a^3} – 3\left(a – \frac{1}{a}\right)\)
Let \(a = 2x\), then:
\((2x – \frac{1}{2x})^3 = (2x)^3 – \frac{1}{(2x)^3} – 3(2x – \frac{1}{2x})\)
Step 2: Substitute the given value:
\(4^3 = 8x^3 – \frac{1}{8x^3} – 3(4)\)
Step 3: Simplify:
\(64 = 8x^3 – \frac{1}{8x^3} – 12\)
\(8x^3 – \frac{1}{8x^3} = 64 + 12 = 76\)
Answer:76

Q18: If \(3x+\frac{1}{3x}=3\), find:

i. \(9x^2 + \frac{1}{9x^2}\)

Step 1: Recall the identity: \((a + \frac{1}{a})^2 = a^2 + \frac{1}{a^2} + 2\)
Step 2: Let \(a = 3x\), then:
\((3x + \frac{1}{3x})^2 = (3x)^2 + \frac{1}{(3x)^2} + 2\)
Step 3: Substitute the given value:
\(3^2 = 9x^2 + \frac{1}{9x^2} + 2\)
Step 4: Solve for \(9x^2 + \frac{1}{9x^2}\):
\(9x^2 + \frac{1}{9x^2} = 9 – 2 = 7\)
Answer:7

ii. \(27x^3 + \frac{1}{27x^3}\)

Step 1: Recall the identity: \((a + \frac{1}{a})^3 = a^3 + \frac{1}{a^3} + 3(a + \frac{1}{a})\)
Step 2: Let \(a = 3x\), then:
\((3x + \frac{1}{3x})^3 = (27x^3 + \frac{1}{27x^3}) + 3(3x + \frac{1}{3x})\)
Step 3: Substitute the given value:
\(3^3 = 27x^3 + \frac{1}{27x^3} + 3(3)\)
Step 4: Simplify:
\(27 = 27x^3 + \frac{1}{27x^3} + 9\)
\(27x^3 + \frac{1}{27x^3} = 27 – 9 = 18\)
Answer:18

Q19: The sum of the square of two numbers is 13 and their product is 6. Find :

i. The sum of the two numbers

Step 1: Let the two numbers be \(a\) and \(b\). Recall the identity:
\((a+b)^2 = a^2 + b^2 + 2ab\)
Step 2: Substitute the given values \(a^2+b^2 = 13\) and \(ab = 6\):
\((a+b)^2 = 13 + 2(6)\)
Step 3: Simplify:
\((a+b)^2 = 13 + 12 = 25\)
\(a+b = \sqrt{25} = \pm 5\)
Answer:\(\pm 5\)

ii. The difference between them.

Step 1: Recall the identity:
\((a-b)^2 = a^2 + b^2 – 2ab\)
Step 2: Substitute the given values:
\((a-b)^2 = 13 – 2(6)\)
Step 3: Simplify:
\((a-b)^2 = 13 – 12 = 1\)
\(a-b = \sqrt{1} = \pm 1\)
Answer:\(\pm 1\)