Question

How many real roots does \[{x^4} - {x^3} - 9{x^2} + 7x + 14 = 0\] have?

Answer 1

Hint: Here we will first use the rational root theorem to find one root of the given equation. Then we will write the equation in the form of the factors such that one factor will be linear polynomial and the other will be cubic polynomial. We will then factorize the cubic equation by taking out common factors to find the remaining factors of the equation. From the factors, we will get the roots of the given equation.

Complete step by step solution:
The given equation is \[{x^4} - {x^3} - 9{x^2} + 7x + 14 = 0\].
We can write this equation as:
\[f\left( x \right) = {x^4} - {x^3} - 9{x^2} + 7x + 14 = 0\]
First, we will use the rational root theorem to find a root of the given equation.
Rational root theorem states that, any rational zeros of a functional equation are expressible in the form \[\dfrac{p}{q}\] for integers \[p\] and \[q\]. Here, \[p\] is the divisor of the constant term of the equation i.e. 14 and \[q\] is the divisor of the coefficient of the leading term i.e. 1.
Therefore the possible rational zeros are \[ \pm 1\], \[ \pm 2\], \[ \pm 7\] and \[ \pm 14\].
So, we will select \[x = - 1\] and put it in the given equation. Therefore, we get
\[f\left( x \right) = {\left( { - 1} \right)^4} - {\left( { - 1} \right)^3} - 9{\left( { - 1} \right)^2} + 7\left( { - 1} \right) + 14\]
Applying the exponent on the terms, we get
\[ \Rightarrow f\left( x \right) = 1 + 1 - 9 - 7 + 14 = 0\]
Hence \[x = - 1\] is the root of the given equation. Therefore, \[\left( {x + 1} \right)\] is the factor of the given equation. Therefore, we can write the equation as
\[ \Rightarrow {x^4} - {x^3} - 9{x^2} + 7x + 14 = \left( {x + 1} \right)\left( {{x^3} - 2{x^2} - 7x + 14} \right)\]
Now we will simplify the cubic factor to get the other roots of the equation. Therefore, we will take \[{x^2}\] common from the first two terms and \[ - 7\] common from the last two terms, we get
\[ \Rightarrow {x^4} - {x^3} - 9{x^2} + 7x + 14 = \left( {x + 1} \right)\left( {{x^2}\left( {x - 2} \right) - 7\left( {x - 2} \right)} \right)\]
Now we will take \[\left( {x - 2} \right)\] common from the second bracket term. Therefore, we get
\[ \Rightarrow {x^4} - {x^3} - 9{x^2} + 7x + 14 = \left( {x + 1} \right)\left( {x - 2} \right)\left( {{x^2} - 7} \right)\]
We can write the above equation as
\[ \Rightarrow {x^4} - {x^3} - 9{x^2} + 7x + 14 = \left( {x + 1} \right)\left( {x - 2} \right)\left( {{x^2} - {{\left( {\sqrt 7 } \right)}^2}} \right)\]
Now we will use the algebraic identity \[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\] in the above equation. Therefore, we get
\[ \Rightarrow {x^4} - {x^3} - 9{x^2} + 7x + 14 = \left( {x + 1} \right)\left( {x - 2} \right)\left( {x + \sqrt 7 } \right)\left( {x - \sqrt 7 } \right)\]

Hence the given equation will have four roots which are \[x = - 1,2,\sqrt 7 , - \sqrt 7 \].

Note:
Here the equation given in the question is the quartic equation in which the highest exponent of the variable \[x\] is four. Roots are those values of the equation where the value of the equation becomes zero. For any equation, numbers of roots are always equal to the value of the highest exponent of the variable \[x\]. For example, a quadratic equation is an equation in which the highest exponent of the variable \[x\] is two and a quadratic equation has only two roots.