Question

If a2 + b2 = 99, ab = 9, find the value of a-b​

Answer 1

$\Large {\underline { \sf{Explication \: of \: Steps :}}}$

$\underline{\pmb { \sf {\maltese \; \; \; Given \: Information \: : \; \; \; }} }$

• a² + b² = 99
• ab = 9

$\underline{\pmb { \sf {\maltese \; \; \; To \: calculate \: : \; \; \; }} }$

• a - b

$\underline{\pmb { \sf {\maltese \; \; \; Calculation \: : \; \; \; }} }$

We know that,

$\bigstar \: \boxed{\sf { {(a-b)}^2= a^2 + b^2 - 2ab }}$

Now from this identity,

$\bigstar \: \boxed{\sf { (a-b)= \sqrt{a^2 + b^2 - 2ab} }}$

• a² + b² = 99
• ab = 9

Substituting values,

→ (a-b) = $\sf {\sqrt{99 - 2(9)} }$

→ (a-b) = $\sf {\sqrt{99 - 18} }$

→ (a-b) = $\sf {\sqrt{81} }$

⠀⠀⠀⠀⠀_____________

By prime factorization,

→ 81 = 9 × 9

• Making pairs and picking out a factor from each pair. Product of those picked factors will be the square root.

→ 81 = 9 × 9

→ √81 = 9

⠀⠀⠀⠀⠀_____________

→ (a-b) = 9

Therefore, value of (a-b) is 9.

Hence, we got the answer !

_______________________________

$\Large {\underline { \sf {More \: Identities :}}}$

$\boxed{\begin{array}{cc}\boxed{\bigstar\:\:\textbf{\textsf{Algebraic\:Identity}}\:\bigstar}\\\\1)\bf\:(A+B)^{2} = A^{2} + 2AB + B^{2}\\\\2)\sf\: (A-B)^{2} = A^{2} - 2AB + B^{2}\\\\3)\bf\: A^{2} - B^{2} = (A+B)(A-B)\\\\4)\sf\: (A+B)^{2} = (A-B)^{2} + 4AB\\\\5)\bf\: (A-B)^{2} = (A+B)^{2} - 4AB\\\\6)\sf\: (A+B)^{3} = A^{3} + 3AB(A+B) + B^{3}\\\\7)\bf\:(A-B)^{3} = A^{3} - 3AB(A-B) + B^{3}\\\\8)\sf\: A^{3} + B^{3} = (A+B)(A^{2} - AB + B^{2})\\\\\end{array}}$