Solveeit Logo
Login

Question

Question: Solve the equation \({x^4} + 2{x^3} - 16{x^2} - 22x + 7 = 0\) given that it has a root \(2 + \sqrt 3...

Solve the equation x4+2x3−16x2−22x+7=0{x^4} + 2{x^3} - 16{x^2} - 22x + 7 = 0 given that it has a root 2+3.2 + \sqrt 3 .

Explanation

Solution

Here in this question one root of the biquadratic equation is given. One thing to notice here is that the root is irrational and we know that irrational roots occur in pairs, that is one root is the conjugate of the other. From this information, we can easily find the other root. We also know the concept of the sum of roots and the product of roots of polynomials. From there we can find the other roots of our equation.

Complete step-by-step solution:
In the given question,
As we know that irrational roots occur in pairs.
So, the other root will be 2−32 - \sqrt 3 .
Now, we have two roots of equation, that is 2+3 and 2−3.2 + \sqrt 3 \,and\,2 - \sqrt 3 .
As we know that
Sum of roots =-coefficient of x3coefficient of x4 = \dfrac{{ \text{-coefficient of}\,{x^3}}}{{\text{coefficient of} \,{x^4}}}
We have,
x4+2x3−16x2−22x+7=0{x^4} + 2{x^3} - 16{x^2} - 22x + 7 = 0
coefficient of x3= 2\text{coefficient of} \,{x^3} = \,2
coefficient of x4=1\text{coefficient of} \,{x^4} = 1
On substituting the values,
Sum of roots = −21 = \,\dfrac{{ - 2}}{1}
Also,
Product of roots =constantcoefficient of x4 = \dfrac{{\text{constant}}}{{\text{coefficient of}\,{x^4}}}
So,
Constant = 7 = \,7
Coefficient of x4=1{x^4} = 1
On substituting the value
Product of roots = 71 = \,\dfrac{7}{1}
Now, let's consider the other two roots be m and n.
Now,
Product of roots = 7 = \,7
(2+3)×(2−3)×m×n=7\left( {2 + \sqrt 3 } \right) \times \left( {2 - \sqrt 3 } \right) \times m \times n = 7
Using, (a+b)(a−b)=a2−b2\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}
((2)2−(3)2)×m×n=7\left( {{{\left( 2 \right)}^2} - {{\left( {\sqrt 3 } \right)}^2}} \right) \times m \times n = 7
⇒(4−3)×m×n=7\Rightarrow \left( {4 - 3} \right) \times m \times n = 7
⇒m×n=7\Rightarrow m \times n = 7
⇒m=7n\Rightarrow m = \dfrac{7}{n}
Now,
Sum of roots = −2 = \, - 2
(2+3)+(2−3)+m+n=−2\left( {2 + \sqrt 3 } \right) + \left( {2 - \sqrt 3 } \right) + m + n = - 2
4+m+n=−24 + m + n = - 2
On substituting the value of m, we get
4+7n+n=−24 + \dfrac{7}{n} + n = - 2
Taking L.C.M
4n+n2+7n=−2\dfrac{{4n + {n^2} + 7}}{n} = - 2
On cross-multiplication, we get
n2+4n+7=−2n{n^2} + 4n + 7 = - 2n
⇒n2+6n+7=0\Rightarrow {n^2} + 6n + 7 = 0
Using formula, x=−b±b2−4ac2ax = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}
On comparing, n2+6n+7=0{n^2} + 6n + 7 = 0 with ax2+bx+c=0a{x^2} + bx + c = 0
a=1  ,  b=6  and c= 7a = 1\,\,,\,\,b = 6\,\,and\,c = \,7
Putting the values in formula
x=−(6)±62−4(1)(7)2(1)x = \dfrac{{ - \left( 6 \right) \pm \sqrt {{6^2} - 4\left( 1 \right)\left( 7 \right)} }}{{2\left( 1 \right)}}
⇒x=−6±36−282\Rightarrow x = \dfrac{{ - 6 \pm \sqrt {36 - 28} }}{2}
⇒x=−6±82\Rightarrow x = \dfrac{{ - 6 \pm \sqrt 8 }}{2}
⇒x=−6±222\Rightarrow x = \dfrac{{ - 6 \pm 2\sqrt 2 }}{2}
⇒x=−3±2\Rightarrow x = - 3 \pm \sqrt 2
Therefore, the other two roots of the equation are −3+2- 3 + \sqrt 2 and −3−2- 3 - \sqrt 2.
Hence, the solutions of above equation are 2+3  ,  2−3  ,  −3+2  and  −3−2.2 + \sqrt 3 \,\,,\,\,2 - \sqrt 3 \,\,,\,\, - 3 + \sqrt 2 \,\,\text{and}\,\, - 3 - \sqrt 2 .

Note: If one root of the equation is irrational then the other root is the conjugate of the first one.
It is a very important property to remember because it is used several times while solving a question of the quadratic or biquadratic equations. Sum of roots and product of roots are also very important concepts of polynomials.