Question

How do you solve the simultaneous equations $y = {x^2} - 2$ and $y = 3x + 8$?

Answer 1

The problem deals with the quadratic equation and a linear equation. A polynomial having degree 2 in one variable in form of $f\left( x \right) = A{x^2} + Bx + C$ where $A,B,C \in R$ and $A \ne 0$, when equated to zero becomes quadratic equation of degree 2 .The general form of quadratic equation is $f\left( x \right) = A{x^2} + Bx + C$. $A$ is called as the leading coefficient and $C$ is called the absolute term of the quadratic equation. The value satisfying the quadratic equation is called the roots of the quadratic equation. The quadratic equation will always have two roots. The nature of the root may be either real or imaginary.
Formula for finding the root of the quadratic equation is $\dfrac{{ - B \pm \sqrt {{B^2} - 4AC} }}{{2A}}$.
Here we will use a substitution method to solve the simultaneous equations. In the substitution method we substitute the value of one variable into the second equation to find the value of the second variable.

Complete step by step answer:
Step: 1 the equation of the given quadratic is $y = {x^2} - 2$.
The given linear equation is $y = 3x + 8$.
Substitute the value of $y = 3x + 8$ in the given quadratic equation $y = {x^2} - 2$.
$\Rightarrow 3x + 8 = {x^2} - 2 \\\ \Rightarrow {x^2} - 2 - 3x - 8 = 0 \\\ \Rightarrow {x^2} - 3x - 10 = 0 \\\$
Step: 2 compare the quadratic equation ${x^2} - 3x - 10 = 0$ with standard form of quadratic equation $f\left( x \right) = a{x^2} + bx + c$.
$\Rightarrow A = 1 \\\ \Rightarrow B = - 3 \\\ \Rightarrow C = - 10 \\\$
Step: 3 substitute the value of $\left( {A,B,C,} \right)$ in the formula of quadratic equation.
$\Rightarrow x = \dfrac{{ - B \pm \sqrt {{B^2} - 4AC} }}{{2A}} \\\ \Rightarrow x = \dfrac{{ - \left( { - 3} \right) \pm \sqrt {{{\left( { - 3} \right)}^2} - 4 \times 1 \times \left( { - 10} \right)} }}{{2 \times 1}} \\\ \Rightarrow x = \dfrac{{3 \pm \sqrt {9 + 40} }}{2} \\\$
Since the quadratic will have two roots so, consider the positive value of $x$.
$x = \dfrac{{3 \pm \sqrt {9 + 40} }}{2} \\\ \Rightarrow x = \dfrac{{3 + \sqrt {49} }}{2} \\\ \Rightarrow x = \dfrac{{3 + 7}}{2} \\\ \Rightarrow x = 5 \\\$
Consider the negative value of $x$.
$x = \dfrac{{3 \pm \sqrt {9 + 40} }}{2} \\\ \Rightarrow x = \dfrac{{3 - \sqrt {49} }}{2} \\\ \Rightarrow x = \dfrac{{3 - 7}}{2} \\\ \Rightarrow x = - 2 \\\$
Therefore, the roots of the equation are $x = 5$ and $x = - 2$.
Step: 4 substitute the value of $x = 5$ in the linear equation $y = 3x + 8$.
$\Rightarrow y = 3x + 8 \\\ \Rightarrow y = 3 \times 5 + 8 \\\ \Rightarrow y = 23 \\\$
Substitute the value of $x = - 2$ in the linear equation $y = 3x + 8$.
$\Rightarrow y = 3x + 8 \\\ \Rightarrow y = 3 \times \left( { - 2} \right) + 8 \\\ \Rightarrow y = - 6 + 8 \\\ \Rightarrow y = 2 \\\$
Final Answer:
Therefore, the solution of the given simultaneous equation is $\left( {5,23} \right)$ and $\left( { - 2,2} \right)$.

Note: Here use the method of substitution to solve the linear equation. Use the quadratic equation formula $\dfrac{{ - B \pm \sqrt {{B^2} - 4AC} }}{{2A}}$ to find the roots of the given quadratic equation. Substitute the roots, $x = - 2$ and $x = 5$ in the quadratic equation one by one in linear equations to get the value of $y$.