Question

4^(x+1)+4^(1-x)=10 solve for x by quadratic formula............... please ​

Answer 1

Given equation :

⇒ 4^( x + 1 ) + 4^( 1 - x ) = 10

Let's recall laws of exponents to be used here

• a^( m + n ) = a^m × a^n
• a^( m - n ) = a^m / a^n

Hence the given equation becomes

$\Rightarrow \sf (4^x \times 4^1) + \dfrac{4^1}{4^x} = 10$

Substituting 4^x = y in the above equation

⇒ 4y + 4/y = 10

Multiplying every term by ' y '

⇒ 4y² + 4 = 10y

⇒ 4y² - 10y + 4 = 0

Dividing throughout the equation by ' 2 '

⇒ 4y²/2 - 10y/2 + 4/2 = 0

⇒ 2y² - 5y + 2 = 0

Comparing 2y² - 5y + 2 with ay² + by + c = 0 we get

• a = 2
• b = - 5
• c = 2

Using Quadratic formula

$\boxed{\rm y = \dfrac{-b \pm \sqrt{ b^2 - 4ac }}{2a} }$

$\Rightarrow \sf y = \dfrac{- ( - 5 ) \pm \sqrt{ ( - 5 )^2 - 4(2)(2) }}{2(2)} } \\\\\\ \Rightarrow \sf y = \dfrac{5 \pm \sqrt{ 25 - 16 }}{4} } \\\\\\ \Rightarrow \sf y = \dfrac{5 \pm \sqrt{9 }}{4} } \\\\\\ \Rightarrow \sf y = \dfrac{5 \pm 3}{4} }$

⇒ y = ( 5 + 3 ) / 4         OR        y = ( 5 - 3 )/ 4

⇒ y = 8/4    OR    y = 2/4

⇒ y = 2       OR    y = 1/2

But y = 4^x

⇒ 4^x = 2     OR       4^x = 1/2

⇒ ( 2² )ˣ = 2      OR      ( 2² )ˣ = 1/2

Since 1 / aⁿ = a⁻ ⁿ  

⇒ 2^( 2x ) = 2¹    OR   2^( 2x ) = 2⁻ ¹

Since Bases are equal we can equate exponents

⇒ 2x = 1      OR        2x = - 1

⇒ x = 1/2     OR        x = - 1/2

∴ the roots of the equations are 1/2 and - 1/2.

Answer 2

Step-by-step explanation:

Answer:

${2}^{x + 1} + {4}^{1 + x} = 10$

${4}^{x} . \frac{4 + 4}{ {4}^{x} } = 10$

${4}^{x} + \frac{1}{ {4}^{x} } = \frac{10}{4} = \frac{5}{2}$

Let,

${4}^{x} = p$

$\frac{p + 1}{p} = \frac{5}{2}$

${2p}^{2} + 2 = 5p$

${2p}^{2} - 5p - 2 = 0$

${2p}^{2} - 4p - p + 2 = 0$

$2p(p - 2) - 1(p - 2) = 0$

$(p - 2)(2p - 1) = 0$

p=2 ,1/2.

$put$

$p = {4}^{x} (or) {2}^{2x}$

${2}^{2x } = 2$

$2x = 1 \: = > x = \frac{1}{2}$

${2}^{2x} = \frac{1}{2} = {2}^{ - 1}$

$2x = - 1 \: = > x = \frac{ - 1}{2}$