Question

Let a, b be the solutions of \[{{x}^{2}}+px+1=0\] and c, d be the solutions of \[{{x}^{2}}+qx+1=0\]. If \[(a-c)(b-c)\] and \[(a+d)(b+d)\] are the solutions of \[{{x}^{2}}+\alpha x+\beta =0\], then \[\beta \] equals
(1) \[p+q\]
(2) \[p-q\]
(3) \[{{p}^{2}}+{{q}^{2}}\]
(4) \[{{q}^{2}}-{{p}^{2}}\]

Answer 1

Hint: Before solving questions of this kind we should know all the formulas related to trigonometry. A binomial equation has two roots and the sum of the roots of the equation is equal to the ratio of the coefficient of x and coefficient of x square and the product of the roots is equal to the ratio of the coefficient of x and coefficient of x square.

Complete step by step answer:
In the above question, it is given that a, b is the solution of the equation \[{{x}^{2}}+px+1\]
So, \[a+b=-p\]
And \[ab=1\]
as we know that,
 \[a+b=\dfrac{-b}{a}\] and
\[ab=\dfrac{c}{a}\]
Also, it is given that c, d is the roots of the equation
\[{{x}^{2}}+qx+1=0\]
So, \[c+d=-q\]
And \[cd=1\]
Now we have to find the value of \[\beta \] in the equation \[{{x}^{2}}+\alpha x+\beta =0\].
And it is given that \[(a-c)(b-c)\] and \[(a+d)(b+d)\] are the roots of the equation \[{{x}^{2}}+\alpha x+\beta =0\].
So, \[(a-c)(b-c)(a+d)(b+d)=\beta \]……….eq(1)
(because the product of roots of the equation will be equal to the constant in the equation)
Now we have the equation
\[(a-c)(b-c)(a+d)(b+d)\]
Now we will multiply them, first, we will multiply \[(a-c)(b-c)\] and then we will multiply \[(a+d)(b+d)\] which is as follows.
\[(a-c)(b-c)(a+d)(b+d)=(ab-ac-bc+{{c}^{2}})(ab+ad+bd+{{d}^{2}})\]
\[\Rightarrow (a-c)(b-c)(a+d)(b+d)=(ab-c(a+b)+{{c}^{2}})(ab+d(a+b)+{{d}^{2}})\]……..eq(2)
We know that \[a+b=-p\]. So we will put this value in eq(2) and the following result will be obtained
\[(a-c)(b-c)(a+d)(b+d)=(ab+pc+{{c}^{2}})(ab-pd+{{d}^{2}})\]
Also, we know that \[ab=1\] now putting this value in the above equation, we get
\[(a-c)(b-c)(a+d)(b+d)=(1+pc+{{c}^{2}})(1-pd+{{d}^{2}})\]
On solving the equation further, we get
\[(a-c)(b-c)(a+d)(b+d)=(1-pd+{{d}^{2}}+pc-{{p}^{2}}cd+pc{{d}^{2}}+{{c}^{2}}-{{c}^{2}}pd+{{c}^{2}}{{d}^{2}})\]…….eq(3)
From the above equations, we know that the value of \[cd=1\]. So on putting this value in eq(3) we get the following result.
\[(a-c)(b-c)(a+d)(b+d)=(1-pd+{{d}^{2}}+pc-{{p}^{2}}+pd+{{c}^{2}}-cp+1)\]
\[\Rightarrow (a-c)(b-c)(a+d)(b+d)=(2-p(c+d)+p(c+d)+{{c}^{2}}+{{d}^{2}}-{{p}^{2}})\]
We know that,
\[\begin{align}
  & c+d=-q \\
 & \\
\end{align}\]
So on putting this value in the above equation we get
\[(a-c)(b-c)(a+d)(b+d)=(2-pq+pq+{{c}^{2}}+{{d}^{2}}-{{p}^{2}})\]…….eq(4)
Also, we know that,
\[{{(c+d)}^{2}}={{c}^{2}}+{{d}^{2}}+2cd\]
\[\Rightarrow {{(c+d)}^{2}}-2cd={{c}^{2}}+{{d}^{2}}\]
On putting this value in eq(4), we get the following results
\[\begin{align}
  & (a-c)(b-c)(a+d)(b+d)=(2-pq+pq+{{(c+d)}^{2}}-2cd-{{p}^{2}}) \\
 & \Rightarrow (a-c)(b-c)(a+d)(b+d)=(2+{{q}^{2}}-2-{{p}^{2}}) \\
 & \Rightarrow (a-c)(b-c)(a+d)(b+d)=(2+{{q}^{2}}-2-{{p}^{2}}) \\
 & \\
\end{align}\]
Both the values of \[2\] and \[pq\]will cancel out each other because the values are the same but the signs of both the values are different, so on solving the equation further we get the following results
\[\begin{align}
  & (a-c)(b-c)(a+d)(b+d)=({{q}^{2}}-{{p}^{2}}) \\
 & \\
\end{align}\] ………eq(5)
And according to our first equation, we know that
\[(a-b)(b-c)(a+d)(b+d)=\beta \]
So from eq(1) and eq(5)
\[\beta ={{q}^{2}}-{{p}^{2}}\]

So, the correct answer is “Option 4”.

Note:
We can also draw the graphs of the equation. If the highest degree of variable in the polynomial is even then both the arms of the graph will be either upwards or downwards and if the highest degree of variable in the polynomial is odd then one arm of the graph will be upwards and one arm will be downwards.