Solveeit Logo
Login

Question

Question: What are the roots of the quadratic equation \({a^2}{b^2}{x^2} - ({a^2} + {b^2})x + 1 = 0\)? 1\. \...

What are the roots of the quadratic equation a2b2x2−(a2+b2)x+1=0{a^2}{b^2}{x^2} - ({a^2} + {b^2})x + 1 = 0?
1. 1a2,1b2\dfrac{1}{{{a^2}}},\dfrac{1}{{{b^2}}}
2. −1a2,−1b2 - \dfrac{1}{{{a^2}}}, - \dfrac{1}{{{b^2}}}
3. 1a2,−1b2\dfrac{1}{{{a^2}}}, - \dfrac{1}{{{b^2}}}
4. −1a2,1b2 - \dfrac{1}{{{a^2}}},\dfrac{1}{{{b^2}}}

Explanation

Solution

We have to find the roots of the equation, which means we have to find the possible solutions of the variable ‘x’. so, we will solve the equation given in the question to find the values of ‘x’. First we will simplify the equation and then solve step by step.

Complete step-by-step solution:
The equation given in the question is:
a2b2x2−(a2+b2)x+1=0{a^2}{b^2}{x^2} - ({a^2} + {b^2})x + 1 = 0
⇒a2b2x2−a2x−b2x+1=0\Rightarrow {a^2}{b^2}{x^2} - {a^2}x - {b^2}x + 1 = 0
⇒(a2b2x2−a2x)−b2x+1=0\Rightarrow ({a^2}{b^2}{x^2} - {a^2}x) - {b^2}x + 1 = 0
We will take a2x{a^2}xcommon from the first two terms.
a2x(b2x−1)−b2x+1=0{a^2}x({b^2}x - 1) - {b^2}x + 1 = 0
Now, we will take the -1 common from the last two terms.
a2x(b2x−1)−1(b2x−1)=0{a^2}x({b^2}x - 1) - 1({b^2}x - 1) = 0
⇒(a2x−1)(b2x−1)=0\Rightarrow ({a^2}x - 1)({b^2}x - 1) = 0
First, we will take the first term to find a value of ‘x’.
a2x−1=0{a^2}x - 1 = 0
⇒a2x=1\Rightarrow {a^2}x = 1
x=1a2x = \dfrac{1}{{{a^2}}}
So, the one value of x is 1a2\dfrac{1}{{{a^2}}}.
Now, we will take the second term to find another value of ‘x’.
⇒b2x−1=0\Rightarrow {b^2}x - 1 = 0
⇒b2x=1\Rightarrow {b^2}x = 1
⇒x=1b2\Rightarrow x = \dfrac{1}{{{b^2}}}
So, the value of x is 1b2\dfrac{1}{{{b^2}}}.
The roots of the equation a2b2x2−(a2+b2)x+1=0{a^2}{b^2}{x^2} - ({a^2} + {b^2})x + 1 = 0 is 1a2,1b2\dfrac{1}{{{a^2}}},\dfrac{1}{{{b^2}}}.
So, option (1) is the correct answer.

Note: Quadratic equations are the polynomial equations of degree 2 in one variable of type f(x)=ax2+bx+cf(x) = a{x^2} + bx + c where a,b,c,∈Ra,b,c, \in R and a≠0a \ne 0 The values of x satisfying the quadratic equation are the roots of the quadratic equation. The quadratic equation will always have two roots. The nature of roots may be either real or imaginary. In other words, x=αx = \alpha is a root of the quadratic equationf(x)f(x), if f(α)=0f(\alpha ) = 0.
The real roots of an equation f(x)=0f(x) = 0are the x-coordinates of the points where the curve y=f(x)y = f(x) intersect the x-axis.

One of the roots of the quadratic equation is zero and the other is −ba\dfrac{{ - b}}{a}if c=0c = 0
Both the roots are zero if b=c=0b = c = 0.
The roots are reciprocal to each other if a=ca = c