Question

quadratic equation class 11​

Answer 1

Question:

Solve for "x" in the given equation;

3^(2x+1) + 3^2 = 3^(x+3) + 3^x

Answer:

x = 2,-1

Note:

• (a^x)•(a^y) = a^(x+y)

• (a^x)/(a^y) = a^(x-y)

• (a^x)^y = a^(x•y)

Solution:

We have;

=> 3^(2x+1) + 3^2 = 3^(x+3) + 3^x

=> (3^2x)•(3^1) + 3^2 = (3^x)•(3^3) + 3^x

=> (3^2x)•3 + 9 = (3^x)•27 + 3^x

=> 3•(3^x)^2 + 9 = (3^x)•(27+1)

=> 3•(3^x)^2 + 9 = 28•(3^x)

Let , 3^x = y

Now, putting y = 3^x in the above equation, we have;

=> 3y^2 + 9 = 28y

=> 3y^2 - 28y + 9 = 0

=> 3y^2 - 27y - y + 9 = 0

=> 3y(y - 9) - (y - 9) = 0

=> (y - 9)(3y - 1) = 0

either (y - 9) = 0 OR (3y - 1) = 0

Case(1), when (y-9) = 0

=> y - 9 = 0

=> y = 9

=> 3^x = 9

=> 3^x = 3^2

=> x = 2

Case(2), when (3y-1) = 0

=> 3y - 1 = 0

=> 3y = 1

=> y = 1/3

=> 3^x = 1/3

=> 3^x = 3^(-1)

=> x = -1

Hence,

The solutions of the given equation are

x = 2 , -1

Answer 2

Answer:

$\large\bold\red{x=2,-1}$

Step-by-step explanation:

It us being Given that,

${3}^{2x + 1} + {3}^{2} = {3}^{x + 3} + {3}^{x}$

But we know that,

${a}^{m + n} = {a}^{m} \times {a}^{n}$

Therefore,

Further simplifying,

we get,

$= > {3}^{2x} \times {3}^{1} + 9 = { 3}^{x} \times {3}^{3} + {3}^{x} \\ \\ = > { ({3}^{x} )}^{2} \times 3 + 9 = 27 \times {3}^{x} + {3}^{x} \\ \\ = > {( {3}^{x}) }^{2} \times 3 + 9 = 28 \times {3}^{x}$

Further,

Let us consider,

${3}^{x} = y$

Therefore,

we get,

$= > 3{y}^{2} + 9 = 28y \\ \\ = > 3 {y}^{2} - 28y + 9 = 0$

Now,

Apply Middle term splitting form,

we get,

$= > 3{y}^{2} - 27y - y + 9 = 0$

Taking common terms,

we get,

$= > 3y(y - 9) - 1(y - 9) = 0 \\ \\ = > (y - 9)(3y - 1) = 0 \\ \\ = > y = 9 \\ \\ and \\ \\ y = \frac{1}{3}$

Therefore,

we get,

${3}^{x} = 9 = {3}^{2} \\ \\ = > x = 2 \\ \\ and \\ \\ {3}^{x} = \frac{1}{3} = {3}^{ - 1} \\ \\ = > x = - 1$

Hence,

x = 2 and x = -1