Question

How do you solve the system $4{{x}^{2}}-56x+9{{y}^{2}}+160=0$ and $4{{x}^{2}}+{{y}^{2}}- 64=0$?

Answer 1

Hint:There are two unknowns $x$ and $y$ and also two quadratic equations to solve. We solve the equation with the process of substitution. We find the value of ${{y}^{2}}$ from one equation and put it on the other equation. Then we solve the quadratic equation. We find the points for the solution thereafter.

Complete step by step solution:
The given equations $4{{x}^{2}}-56x+9{{y}^{2}}+160=0$ and $4{{x}^{2}}+{{y}^{2}}-64=0$ are quadratic equations of two variables.We know that the number of equations has to be equal to the number of unknowns to solve them.We the value of ${{y}^{2}}$ from the equation $4{{x}^{2}}+{{y}^{2}}-64=0$ where ${{y}^{2}}=64-
4{{x}^{2}}$. We replace the value of ${{y}^{2}}$ in the second equation of $4{{x}^{2}}-
56x+9{{y}^{2}}+160=0$ and we get
\[\begin{align}
& 4{{x}^{2}}-56x+9{{y}^{2}}+160=0 \\
& \Rightarrow 4{{x}^{2}}-56x+9\left( 64-4{{x}^{2}} \right)+160=0 \\
& \Rightarrow 4{{x}^{2}}-56x+576-36{{x}^{2}}+160=0 \\
& \Rightarrow 32{{x}^{2}}+56x-736=0 \\
& \Rightarrow 4{{x}^{2}}+7x-92=0 \\
\end{align}\]

We get the quadratic equation of $x$ as \[4{{x}^{2}}+7x-92=0\] and solve to get
\[x=\dfrac{-7\pm \sqrt{{{7}^{2}}-4\times \left( -92 \right)\times 4}}{2\times 4}\\
\Rightarrow x =\dfrac{-7\pm\sqrt{1521}}{8}\\
\Rightarrow x =\dfrac{-7\pm 39}{8}\\
\Rightarrow x =-\dfrac{23}{4},4\]
Putting the value of \[x=-\dfrac{23}{4}\] in ${{y}^{2}}=64-4{{x}^{2}}$, we get
\[{{y}^{2}}=64 - 4{{\left( -\dfrac{23}{4} \right)}^{2}}\\
\Rightarrow{{y}^{2}} =64-\dfrac{529}{4}\\
\Rightarrow{{y}^{2}} =\dfrac{-273}{4}\]
This is not possible as square value can’t be negative. Therefore, \[x=-\dfrac{23}{4}\] is not possible. Putting the value of \[x=4\] in ${{y}^{2}}=64-4{{x}^{2}}$, we get \[{{y}^{2}}=64-4{{\left( 4\right)}^{2}}=0\] which gives $y=0$.

Therefore, the solution is \[x=4,y=0\].

Note: We know for a general equation of quadratic $a{{x}^{2}}+bx+c=0$, the value of the roots of $x$ will be $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. This is the quadratic equation solving method. The root part $\sqrt{{{b}^{2}}-4ac}$ of $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ is called the discriminant of the equation. In the given equation we have \[4{{x}^{2}}+7x-92=0\]. The values of a, b, c is $4,7,-92$ respectively.