Exponential Equations in Quadratic Form: Equation, Type, Questions

Exponential Equations in Quadratic form

A polynomial equation in which the highest degree of a variable term is 2 is called a quadratic equation.

Standard form of quadratic equation is $a x^2+b x+c=0$

Where a, b, and c are constants (they may be real or imaginary) and called the coefficients of the equation and $a \neq 0$ (a is also called the leading coefficient).

Eg, $-5 x^2-3 x+2=0, x^2=0,(1+i) x^2-3 x+2 i=0$

As the degree of the quadratic polynomial is 2, so it always has 2 roots (number of real roots + number of imaginary roots = 2)

Roots of quadratic equation

The root of the quadratic equation is given by the formula:

$\begin{aligned} & x=\frac{-b \pm \sqrt{D}}{2 a} \\ & \text { or } \\ & x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\end{aligned}$

Where D is called the discriminant of the quadratic equation, given by $D=b^2-4 a c$,

An equation of the form a^x = b is known as an exponential equation, where

(i) $\mathrm{x} \in \phi$, if $\mathrm{b} \leq 0$
(ii) $x=\log _a b$, if $b>0, a \neq 0$
(iii) $\mathrm{x} \in \phi$, if $\mathrm{a}=1, \mathrm{~b} \neq 1$
(iv) $\mathrm{x} \in \mathrm{R}$, if $\mathrm{a}=1, \mathrm{~b}=1$ (since $1^{\mathrm{x}}=1 \Rightarrow 1=1, \mathrm{x} \in \mathrm{R}$ )

Some special cases of the exponential equation

1. Equation of the form $a^{f(x)}=1$, where $a>0$ and $a \neq 1$, then solve $f(x)=0$

For Example

The given equation is $7^{x^2+4 x+4}=1$
$\Rightarrow \mathrm{x}^2+4 \mathrm{x}+4=0 \quad\left[\because \mathrm{a}^0=1, \mathrm{a}\right.$ is constant $]$
$\Rightarrow(\mathrm{x}+2)(\mathrm{x}+2)=0 \Rightarrow \mathrm{x}=-2$

2. Equation of the form $f\left(a^x\right)=0$ then $f(t)=0$ where $t=a^x$

For example

The given equation is $4^x-3 \cdot 2^x-4=0$ equation is quadratic in $2^x$, so substitute $2^x=t$

$
\begin{aligned}
& \Rightarrow\left(2^{\mathrm{x}}\right)^2-3\left(2^{\mathrm{x}}\right)-4=0 \\
& \Rightarrow \mathrm{t}^2-3 \mathrm{t}-4=0 \\
& \Rightarrow(\mathrm{t}-4)(\mathrm{t}+1)=0 \\
& \Rightarrow \mathrm{t}=4, \mathrm{t}=-1
\end{aligned}
$

since, $\mathrm{t}=2^{\mathrm{x}}$

$
2^x=4 \Rightarrow 2^x=2^2 \Rightarrow x=2
$

and, $2^{\mathrm{x}}=-1$, No solution

Finally we get $x=2$

Summary

Exponential equations in quadratic form present an intriguing intersection of exponential and quadratic functions. By transforming exponential equations into quadratic form, we can apply familiar quadratic-solving techniques to find solutions. Understanding and mastering these transformations expand our ability to tackle a broader range of mathematical problems, enhancing both theoretical knowledge and practical problem-solving skills.

Recommended Video Based on Exponential Equations in Quadratic Form:

Solved Examples Based on Exponential Equations in Quadratic form:

Example 1: If the sum of all the roots of the equation $\mathrm{e}^{2 x}-11 \mathrm{e}^x-45 \mathrm{e}^{-x}+\frac{81}{2}=0$ is $\log _{\mathrm{e}} \mathrm{p}$, then $\mathrm{p}$ is equal to______________.

1) 45

2) 17

3) 46

4) 16

Solution

The equation can be rewritten as

$2 e^{3 x}-22 e^{2 x}+81 e^x-90=0$

Let the roots are $\alpha, \beta, \gamma$

Let $\mathrm{e}^{\mathrm{x}}=\mathrm{t}_{,} \mathrm{e}^\alpha=\mathrm{t}_1, \mathrm{e}^\beta=\mathrm{t}_2, \mathrm{e}^\gamma=\mathrm{t}_3$

$2 t^3-22 t^2+81 t-90=0$

Product of roots $=\mathrm{t}, \mathrm{t}_2 \mathrm{t}_3=\mathrm{e}^\alpha \cdot \mathrm{e}^\beta \cdot \mathrm{e}^\gamma=\frac{90}{2}=45$

$\Rightarrow \mathrm{e}^{\alpha+\beta+\gamma}=45$

$\Rightarrow \alpha+\beta+\gamma=\log _e 45$

So $\mathrm{p}=45$

Hence, the answer is 45.

Example 2: The mean and standard deviation of 15 observations are found to be 8 and 3 respectively. On rechecking, it was found that, in the observations, 20 was misread as 5 . Then, the correct variance is equal to_______.

1) 17

2) 43

3) 81

4) 12

Solution

$\frac{\displaystyle\sum_{i=1}^{14} x i+5}{15}=8$
$\Rightarrow \displaystyle\sum_{i=1}^{14} x i=115 \Rightarrow \displaystyle\sum_{i=1}^{11} x i+20=123$
$\Rightarrow$ Real Mean $=\frac{\displaystyle\sum_{i=1}^{14} x i+20}{15}=\frac{13}{15}$
$\frac{\displaystyle\sum_{i=1}^{14} x i^2+5^2}{15}-(8)^2=9 \Rightarrow \displaystyle\sum_{i=1}^{14} x i^2=1070$
$\begin{aligned} \text { So Real variance } & =\frac{\displaystyle\sum_{i=1}^{14} \times i^2+20^2}{15}-(9)^2 \\ & =\frac{1070+400}{15}-(9)^2 \\ & =\frac{1470}{15}-81\end{aligned}$
$=98-81=17$

Hence, the answer is 17.

Example 3: The number of real roots of the equation $e^{6 x}-e^{4 x}-2 e^{3 x}-12 e^{2 x}+e^x+1=0$ is

1) 2

2) 4

3) 6

4) 1

Solution
$
\begin{aligned}
& e^{6 x}-e^{4 x}-2 e^{3 x}-12 e^{2 x}+e^x+1=0 \\
& \left(\left(c^{3 x}\right)^2-2 e^{3 x}+1\right)-c^x\left(e^{3 x}-1\right)-12 e^{2 r}=0 \\
& \Rightarrow\left(e^{3 x}-1\right)^2-e^x\left(e^{3 x}-1\right)-12\left(e^x\right)^2=0 \\
& \text { Let } e^{3 r}-1=p \text { and } e^2=q \\
& =p^2-p q-12 q^2=0 \\
& \Rightarrow p^2-4 p q+3 p q-12 q^2=0 \\
& B(p-4 q)(p+3 q)=0 \\
& E p=4 q \text { or } p=-3 q \\
& \Rightarrow e^{3 x}-1=4 e^x \text { or } e^{3 x}-1=-3 e^x \\
& \operatorname{Let} e^x=t(\therefore t>0) \\
& \Rightarrow t^3-4 t-1=0 \text { or } t^3+3 t-1=0 \\
& \operatorname{Let} f(t)=t^3-4 t-1 \& \operatorname{Let} g(t)=t^3+3 t-1 \\
& f^{\prime}(t)=3 t^2-4 \& g^{\prime}(t)=3 t^2+3
\end{aligned}
$

$f(t)$ has one positive root $\mathrm{t}=1 \& g(t)$ has one positive root (say $\mathrm{t}_1$ )
So 2 solutions

Example 4: The number of real roots of the equation $e^{4 r}-e^{3 r}-4 e^{2 r}-e^r+1=0$ is equal to $\qquad$.

1) 2

2) 0

3) 1

4) 3

Solution

$
e^{4 x}-e^{2 x}-4 e^{2 x}-e^x+1=0
$
Let $e^x=1$

$
\begin{aligned}
& t^1-t^4-4 t^2-t+1=0 \\
& \Rightarrow t^2-t-4-\frac{1}{t}+\frac{1}{t^2}=0 \\
& \Rightarrow\left(t^2+\frac{1}{t^2}\right)-\left(t+\frac{1}{t}\right)-4=0 \\
& \Rightarrow\left(t+\frac{1}{t}\right)^2-\left(t+\frac{1}{t}\right)-6=0
\end{aligned}
$
Let $t+\frac{1}{t}=t$

$
\begin{aligned}
& \Rightarrow u^2-u-6=0 \\
& \Rightarrow(u-3)(u+2)=0 \\
& \Rightarrow u=3,-2 \\
& \Rightarrow t+\frac{1}{t}=3 \quad\left(\text { As } t+\frac{1}{t}=c^x+\frac{1}{c^x}>0\right) \\
& \Rightarrow t^2-3 t+1=0 \\
& \Rightarrow t=\frac{3 \pm \sqrt{9-4}}{2}=\frac{3+\sqrt{5}}{2}, \frac{3-\sqrt{5}}{2} \\
& \Rightarrow e^x=\frac{3+\sqrt{5}}{2}, e^t=\frac{3-\sqrt{5}}{2} \\
& \Rightarrow x=\ln \left(\frac{3+\sqrt{5}}{2}\right) \cdot \ln \left(\frac{3-\sqrt{5}}{2}\right)
\end{aligned}
$
Hence, the answer is the option 1.

Example 5: The number of real roots of the equation $\mathrm{e}^{4 x}+2 \mathrm{e}^{3 x}-\mathrm{e}^x-6=\mathrm{n}$ is:

1) 0

2) 1

3) 4

4) 2

Solution

$f(x)=e^{4 x}+2 e^{3 x}-e^x-6$
$f^{\prime}(x)=4 e^{4 x}+6 e^{3 x}-e^x=0$$\begin{aligned} & e^x\left(4 e^{3 x}+6 e^{2 x}-1\right)=0 \\ & e^x>0 ; g(x)=4 e^{3 x}+6 e^{2 x}-1 \\ & a(-\infty)=-1: g(\infty)=\infty\end{aligned}$
and $g(x)$ is always increasing function
$\Rightarrow g(x)$ will have exactly one real root
$\Rightarrow f^{\prime}(x)$ will have exactly one real root which will be point of minima
Also $f(0)=-4$$\begin{aligned} & f(-\infty)=-6 \\ & f(\infty)=\infty\end{aligned}$

$\Rightarrow f(x)=0$ will have exactly 1 real root.