Question

Find the number of real solutions of the equation $x^4+8x^2+16=4x^2-12x+9$.
A. 1
B. 2
C. 3
D. 4

Answer 1

Hint: We need to form the given equation into the square form on both sides of the equations. Then we need to solve two quadratic equations separately to find the four possible solutions. At last we need to check the solutions being real.

Complete step-by-step answer:
The given equation is $x^4+8x^2+16=4x^2-12x+9$.
We need to form the squares on both sides. We know ${(a+b)}^2=a^2+b^2+2ab$.
For left hand side $x^4+8x^2+16={\left(x^2\right)}^2+2\times 4\times \left(x^2\right)+4^2={(x^2+4)}^2$.
For the right side $4x^2-12x+9={\left(2x\right)}^2-2\times 3\times \left(2x\right)+3^2={(2x-3)}^2$.
We equate these two to find ${(x^2+4)}^2={(2x-3)}^2$.
Now we square root both sides and get $$(x^2+4)=\pm (2x-3)$$.
As we are taking square root, we have to take both signs of the square root.
So, we got two equations $$(x^2+4)=(2x-3)$$, $$(x^2+4)=-(2x-3)$$.
We need to solve both equations.
For the first equation we have $$(x^2+4)=(2x-3)\Rightarrow x^2-2x+7=0$$
We also know that for general equation of $$a{x}^2+bx+c=0$$, if we find that the discriminant $D=b^2-4ac$ is less than 0 then we have only imaginary solutions of the quadratic equation.
In case of $$x^2-2x+7=0$$, we find $D=b^2-4ac={(-2)}^2-4\times 1\times 7=4-28=-24<0$.
So, we will have two imaginary solutions of $$x^2-2x+7=0$$.
For the second equation we have $$(x^2+4)=-(2x-3)\Rightarrow x^2+2x+1=0$$
We can convert it into square form which is
$$\begin{array}{rl}& x^2+2x+1=0\\ & \Rightarrow {(x+1)}^2=0\end{array}$$
We find two solutions of the equations which are similar in value $x=-1$.
So, the number of real solutions of the equation $x^4+8x^2+16=4x^2-12x+9$ is only 1.

So, the correct answer is “Option A”.

Note: We need to find the solution for only real. No imaginary solutions will be considered. From the discriminant value of the second equation we can tell that the solutions’ value will be the same as if the discriminant value $D=b^2-4ac$ is equal to 0 then we have equal solutions of the quadratic equation. For $$x^2+2x+1=0$$, $D=b^2-4ac=2^2-4\times 1\times 1=4-4=0$.