Question
Solve the following quadratic equation:
$4^{(x+1)} + 4^{(1-x)} = 10$
$4^{(x+1)} + 4^{(1-x)} = 10$
✓
Answer
$4^{(x+1)} + 4^{(1-x)} = 10$
$4^x.4^1 + 4^1.4^{-x} = 10$
$\Rightarrow\text{4y}+\frac{4}{\text{y}}=10$ where $4^x = y$
$\Rightarrow 4y^2 - 10y + 4 = 0$
$\Rightarrow 4y^2 - 8y - 2y + 4 = 0$
$\Rightarrow 4y(y - 2) - 2(y - 2) = 0$
$\Rightarrow (y - 2)(4y - 2) = 0$
⇒ y - 2 = 0 or 4y - 2 = 0
⇒ y = 2 or $\text{y}=\frac{2}{4}=\frac{1}2{}$
⇒ y = 2 or $\text{y}=\frac{1}{2}$
In case I:
$\Rightarrow 4^x = 2$
$\Rightarrow (2)^{2x} = (2)^1$
$\Rightarrow 2x = 1$
$\Rightarrow\text{x}=\frac{1}{2}$
In case II:
$\Rightarrow4^\text{x}=\frac{1}{2}$
$\Rightarrow(2)^\text{2x}=\Big(\frac{1}{2}\Big)^1$
$\Rightarrow(2)^{\text{2x}}=(2)^{-1}$
$\Rightarrow\text{x}=-\frac{1}{2}$
Hence, $\frac{1}{2},-\frac{1}2{}$ are the roots of given equation.
$4^x.4^1 + 4^1.4^{-x} = 10$
$\Rightarrow\text{4y}+\frac{4}{\text{y}}=10$ where $4^x = y$
$\Rightarrow 4y^2 - 10y + 4 = 0$
$\Rightarrow 4y^2 - 8y - 2y + 4 = 0$
$\Rightarrow 4y(y - 2) - 2(y - 2) = 0$
$\Rightarrow (y - 2)(4y - 2) = 0$
⇒ y - 2 = 0 or 4y - 2 = 0
⇒ y = 2 or $\text{y}=\frac{2}{4}=\frac{1}2{}$
⇒ y = 2 or $\text{y}=\frac{1}{2}$
In case I:
$\Rightarrow 4^x = 2$
$\Rightarrow (2)^{2x} = (2)^1$
$\Rightarrow 2x = 1$
$\Rightarrow\text{x}=\frac{1}{2}$
In case II:
$\Rightarrow4^\text{x}=\frac{1}{2}$
$\Rightarrow(2)^\text{2x}=\Big(\frac{1}{2}\Big)^1$
$\Rightarrow(2)^{\text{2x}}=(2)^{-1}$
$\Rightarrow\text{x}=-\frac{1}{2}$
Hence, $\frac{1}{2},-\frac{1}2{}$ are the roots of given equation.
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
From the top of a 9 m high building, the angle of elevation of the top of a cable tower is $60^{\circ}$ and angle of depression of its foot is $45^{\circ}$. Determine the height of the tower and distance between building and tower. $($ Use $\sqrt{3}=1.732)$
→A wire is bent to form a square enclosing an area of $484m^2$. Using the same wire, a circle is formed. Find the area of the circle.
→If 1 and -2 are two zeros of the polynomial ($x^3 - 4x^2 - 7x + 10)$, find its third zero.
→Solve the following quadratic equations by factorization:
$\frac{\text{a}}{\text{x}-\text{b}}+\frac{\text{b}}{\text{x}-\text{a}}=2$
→$\frac{\text{a}}{\text{x}-\text{b}}+\frac{\text{b}}{\text{x}-\text{a}}=2$
Divide a line segment of length 9cm internally in the ratio 4 : 3. Also, give justification of the construction.
Find the roots of the following equations, if they exist, by applying the quadratic formula:$\frac{\text{m}}{\text{n}}\text{x}^2+\frac{\text{n}}{\text{m}}=1-2\text{x}$
→During a medical check-up, the number of heartbeats per minute of 30 patients were recorded and summarised as follows:
Mean, median, mode of grouped data, cumulative frequency graph and ogive.
Find the mean heartbeats per minute for these patients, choosing a suitable method.
| Number of heartbeats per minute | 65-68 | 68-71 | 71-74 | 74-77 | 77-80 | 80-83 | 83-86 |
| Number of patients | 2 | 4 | 3 | 8 | 7 | 4 | 2 |
Find the mean heartbeats per minute for these patients, choosing a suitable method.
In a hospital, the ages of diabetic patients were recorded as follows. Find the median age.
| Age (in years) | 0-15 | 15-30 | 30-45 | 45-60 | 60-75 |
| Number of patients | 5 | 20 | 40 | 50 | 25 |
A wooden toy is in the shape of a cone mounted on a cylinder, as shown in the figure. The total height of the toy is 26cm, while the height of the conical part is 6cm. The diameter of the base of the conical part is 5cm and that of the cyliridrical part is 4cm. The conical part and the cylindrical part are respectively painted red and white. Find the area to be painted by each of these colours. $\big[$Take $\pi=\frac{22}{7}\big]$
→