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Solving Quadratic Systems

Solving Quadratic Systems 1.5 - Solution

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We want to solve the given system of equations using the Substitution Method. {y=x23x3(I)y=-2x2x+5(II)\begin{cases}y=x^2-3x-3 & \, \text {(I)}\\ y=\text{-}2x^2-x+5 & \text {(II)}\end{cases} The y-y\text{-}variable is isolated in Equation (II). This allows us to substitute its value, -2x2x+5,\text{-}2x^2-x+5, for yy in Equation (I).
{y=x23x3(I)y=-2x2x+5(II)\begin{cases}y=x^2-3x-3 & \, \text {(I)}\\ y=\text{-}2x^2-x+5 & \text {(II)}\end{cases}
{-2x2x+5=x23x3y=-2x2x+5\begin{cases}{\color{#0000FF}{\text{-}2x^2-x+5}}=x^2-3x-3 \\ y=\text{-}2x^2-x+5 \end{cases}
(I): {\color{#8C8C8C}{\text{(I): }}} Simplify
{0=x23x3(-2x2x+5)y=-2x2x+5\begin{cases}0=x^2-3x-3-(\text{-}2x^2-x+5) \\ y=\text{-}2x^2-x+5 \end{cases}
{0=x23x3+2x2+x5y=-2x2x+5\begin{cases}0=x^2-3x-3+2x^2+x-5 \\ y=\text{-}2x^2-x+5 \end{cases}
{0=3x22x8y=-2x2x+5\begin{cases}0=3x^2-2x-8 \\ y=\text{-}2x^2-x+5 \end{cases}
{3x22x8=0y=-2x2x+5\begin{cases}3x^2-2x-8=0 \\ y=\text{-}2x^2-x+5 \end{cases}
Notice that in Equation (I), we have a quadratic equation in terms of only the x-x\text{-}variable. 3x22x8=03x2+(-2)x+(-8)=0\begin{gathered} 3x^2-2x-8=0 \quad \Leftrightarrow \quad {\color{#0000FF}{3}}x^2+({\color{#FF0000}{\text{-} 2}})x+({\color{#009600}{\text{-} 8}})=0 \end{gathered} Now, recall the Quadratic Formula. x=-b±b24ac2a\begin{gathered} x=\dfrac{\text{-} b\pm\sqrt{b^2-4ac}}{2a} \end{gathered} We can substitute a=3,{\color{#0000FF}{a}}={\color{#0000FF}{3}}, b=-2,{\color{#FF0000}{b}}={\color{#FF0000}{\text{-} 2}}, and c=-8{\color{#009600}{c}}={\color{#009600}{\text{-} 8}} into this formula to solve the quadratic equation.
x=-b±b24ac2ax=\dfrac{\text{-} b\pm\sqrt{b^2-4ac}}{2a}
x=-(-2)±(-2)24(3)(-8)2(3)x=\dfrac{\text{-} ({\color{#FF0000}{\text{-} 2}})\pm\sqrt{({\color{#FF0000}{\text{-} 2}})^2-4({\color{#0000FF}{3}})({\color{#009600}{\text{-} 8}})}}{2({\color{#0000FF}{3}})}
Solve for xx
x=2±(-2)24(3)(-8)2(3)x=\dfrac{2\pm\sqrt{(\text{-}2)^2-4(3)(\text{-}8)}}{2(3)}
x=2±44(3)(-8)2(3)x=\dfrac{2\pm\sqrt{4-4(3)(\text{-}8)}}{2(3)}
x=2±4+96)6x=\dfrac{2\pm\sqrt{4+96)}}{6}
x=2±1006x=\dfrac{2\pm\sqrt{100}}{6}
x=2±106x=\dfrac{2 \pm 10}{6}

This result tells us that we have two solutions for x.x. One of them will use the positive sign and the other one will use the negative sign.

x=2±106x=\dfrac{2 \pm 10}{6}
x1=2+106x_1=\dfrac{2+10}{6} x2=2106x_2=\dfrac{2-10}{6}
x1=126x_1=\dfrac{12}{6} x2=-86x_2=\dfrac{\text{-} 8}{6}
x1=2x_1=2 x2=-43x_2=\text{-} \dfrac{4}{3}
Now, consider Equation (II). y=-2x2x+5\begin{gathered} y=\text{-}2 x^2 - x + 5 \end{gathered} We can substitute x=2x=2 and x=-43x=\text{-} \frac{4}{3} into the above equation to find the values for y.y. Let's start with x=2.x=2.
y=-2x2x+5y=\text{-}2 x^2-x+5
y=-2(2)22+5y=\text{-}2 ({\color{#0000FF}{2}})^2-{\color{#0000FF}{2}}+5
Solve for yy
y=-2(4)2+5y=\text{-}2(4)-2+5
y=-82+5y=\text{-}8-2+5
y=-5y=\text{-}5
We found that y=-5,y=\text{-} 5, when x=2.x=2. One solution of the system, which is a point of intersection of the two parabolas, is (2,-5).(2,\text{-} 5). To find the other solution, we will substitute -43\text{-} \frac{4}{3} for xx in Equation (II) again.
y=-2x2x+5y=\text{-}2x^2-x+5
y=-2(-43)2(-43)+5y=\text{-}2\left( {\color{#0000FF}{\text{-} \dfrac{4}{3}}} \right)^2-\left({\color{#0000FF}{\text{-}\dfrac{4}{3}}}\right)+5
Solve for yy
y=-2(169)(-43)+5y=\text{-}2\left(\dfrac{16}{9}\right)-\left(\text{-}\dfrac{4}{3}\right)+5
y=-329(-43)+5y=\text{-}\dfrac{32}{9}-\left(\text{-}\dfrac{4}{3}\right)+5
y=-329+43+5y=\text{-}\dfrac{32}{9}+\dfrac{4}{3}+5
y=-329+129+5y=\text{-}\dfrac{32}{9}+\dfrac{12}{9}+5
y=-329+129+459y=\text{-}\dfrac{32}{9}+\dfrac{12}{9}+\dfrac{45}{9}
y=259y=\dfrac{25}{9}
We found that y=259,y=\frac{25}{9}, when x=-43.x=\text{-} \frac{4}{3}. Therefore, our second solution, which is the other point of intersection of the two parabolas, is (-43,259).\left( \text{-} \frac{4}{3},\frac{25}{9}\right).