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Question: How do you solve the system \(4{{x}^{2}}-56x+9{{y}^{2}}+160=0\) and \(4{{x}^{2}}+{{y}^{2}}- 64=0\)?...

How do you solve the system 4x2−56x+9y2+160=04{{x}^{2}}-56x+9{{y}^{2}}+160=0 and 4x2+y2−64=04{{x}^{2}}+{{y}^{2}}- 64=0?

Explanation

Solution

There are two unknowns xx and yy and also two quadratic equations to solve. We solve the equation with the process of substitution. We find the value of y2{{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 4x2−56x+9y2+160=04{{x}^{2}}-56x+9{{y}^{2}}+160=0 and 4x2+y2−64=04{{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 y2{{y}^{2}} from the equation 4x2+y2−64=04{{x}^{2}}+{{y}^{2}}-64=0 where y2=64−4x2{{y}^{2}}=64- 4{{x}^{2}}. We replace the value of y2{{y}^{2}} in the second equation of 4x2−56x+9y2+160=04{{x}^{2}}- 56x+9{{y}^{2}}+160=0 and we get

& 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{aligned}$$ We get the quadratic equation of $x$ as $$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.