Question

If 12(9^x)-35(6^x)+18(4^x)=0 then sum of real values of x satisfying the given equation


fast...​

Answer 1

Answer:

Basic Algebra Identities

(a + b)2 = a2 + b2 + 2ab

(a – b)2 = a2 + b2 – 2ab

a2 – b2 = (a + b)(a – b)

a2 + b2 = (a + b)2 – 2ab = (a – b)2 + 2ab

a3 + b3 = (a + b)(a2 – ab + b2)

a3 – b3 = (a – b)(a2 + ab + b2)

(a + b)3 = a3 + 3ab(a + b) + b3

(a – b)3 = a3 – 3ab(a – b) – b3

Answer 2

The sum of real values of x satisfying the given equation = 1

Given :

$\displaystyle \sf 12( {9}^{x} ) - 35( {6}^{x} ) + 18( {4}^{x} ) = 0$

To find :

The sum of real values of x satisfying the given equation

Solution :

Step 1 of 3 :

Write down the given equation

Here the given equation is

$\displaystyle \sf 12( {9}^{x} ) - 35( {6}^{x} ) + 18( {4}^{x} ) = 0$

Step 2 of 3 :

Find all real values of x satisfying the given equation

$\displaystyle \sf 12( {9}^{x} ) - 35( {6}^{x} ) + 18( {4}^{x} ) = 0$

$\displaystyle \sf{ \implies }12 \big( {( {3}^{2} )}^{x} \big)- 35\big( {(3 \times 2)}^{x} \big) + 18\big( {( {2}^{2} )}^{x} \big) = 0$

$\displaystyle \sf{ \implies }12( {3}^{2x} ) - 35( {3}^{x} \times {2}^{x} ) + 18( {2}^{2x} ) = 0$

$\displaystyle \sf Let \: \: a = {3}^{x} \: \: and \: \: b = {2}^{x}$

Then above equation becomes

$\displaystyle \sf 12 {a}^{2} - 35ab + 18 {b}^{2} = 0$

$\displaystyle \sf{ \implies } 12 {a}^{2} - (8 + 27)ab + 18 {b}^{2} = 0$

$\displaystyle \sf{ \implies } 12 {a}^{2} - 8 ab - 27ab + 18 {b}^{2} = 0$

$\displaystyle \sf{ \implies }4a(3a - 2b) - 9b(3a - 2b) = 0$

$\displaystyle \sf{ \implies }(3a - 2b) (4a - 9b) = 0$

$\displaystyle \sf \therefore Either \: \: (3a - 2b) = 0 \: \: \: or \: \: \: (4a - 9b) = 0$

Now,

$\displaystyle \sf (3a - 2b) = 0 \: \: \: gives$

$\displaystyle \sf 3a = 2b$

$\displaystyle \sf{ \implies } \frac{a}{b} = \frac{2}{3}$

$\displaystyle \sf{ \implies } \frac{ {3}^{x} }{ {2}^{x} } = \frac{2}{3}$

$\displaystyle \sf{ \implies } {\bigg( \frac{3}{2} \bigg)}^{x} = {\bigg( \frac{3}{2} \bigg)}^{ - 1}$

$\displaystyle \sf{ \implies }x = - 1$

Again,

$\displaystyle \sf (4a - 9b) = 0 \: \: \: gives$

$\displaystyle \sf 4a = 9b$

$\displaystyle \sf{ \implies } \frac{a}{b} = \frac{9}{4}$

$\displaystyle \sf{ \implies } \frac{ {3}^{x} }{ {2}^{x} } = \frac{9}{4}$

$\displaystyle \sf{ \implies } {\bigg( \frac{3}{2} \bigg)}^{x} = {\bigg( \frac{3}{2} \bigg)}^{ 2}$

$\displaystyle \sf{ \implies }x = 2$

So the real values of x satisfying the given equation are - 1 and 2

Step 3 of 3 :

Find sum of real values of x satisfying the given equation

The real values of x satisfying the given equation are - 1 and 2

∴ The sum of real values of x satisfying the given equation

= - 1 + 2

= 1

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