# What is the sum of the roots of the equation 4^x - 3(2^(x+3)) + 128 = 0?

Nov 29, 2016

Given equation

${4}^{x} - 3 \left({2}^{x + 3}\right) + 128 = 0$

$\implies {\left({2}^{2}\right)}^{x} - 3 \left({2}^{x} \cdot {2}^{3}\right) + 128 = 0$

$\implies {\left({2}^{x}\right)}^{2} - 3 \left({2}^{x} \cdot 8\right) + 128 = 0$

Taking ${2}^{x} = y$ the equation becomes

$\implies {y}^{2} - 24 y + 128 = 0$

$\implies {y}^{2} - 16 y - 8 y + 128 = 0$

$\implies y \left(y - 16\right) - 8 \left(y - 16\right) = 0$

$\implies \left(y - 16\right) \left(y - 8\right) = 0$

So $y = 8 \mathmr{and} y = 16$

when $y = 8 \implies {2}^{x} = {2}^{3} \implies x = 3$

when $y = 16 \implies {2}^{x} = {2}^{4} \implies x = 4$

Hence roots are $3 \mathmr{and} 4$

So the sum of the roots is $= 3 + 4 = 7$

Nov 29, 2016

$7$

#### Explanation:

If $p \left(x\right) = \left(x - a\right) \left(x - b\right) = {x}^{2} - \left(a + b\right) x + a b$

the $x$ coefficient is the sum of roots.

In ${\left({2}^{x}\right)}^{2} - 24 \cdot {2}^{x} + 128$ we have that

$24$ is the sum of ${r}_{1}$ and ${r}_{2}$ such that

$\left({2}^{x} - {r}_{1}\right) \left({2}^{x} - {r}_{2}\right) = 0$

Also we have ${r}_{1} {r}_{2} = {2}^{7} = {2}^{3} {2}^{4}$ and
${r}_{1} + {r}_{2} = 3 \cdot {2}^{3} = {2}^{3} + {2}^{4}$

then

${r}_{1} = {2}^{3} \to {x}_{1} = 3$ and
${r}_{2} = {2}^{4} \to {x}_{2} = 4$ so

${x}_{1} + {x}_{2} = 7$