3C - A line and a quadratic together

3C - A line and a quadratic together

Can two equations describe the same point in more than one way? You will substitute a linear relation into a quadratic equation, find every matching coordinate pair and check both equations. You will then connect two, one or no real pairs to intersections and tangency.

Reducing to a quadratic

Which points satisfy both y=x+1y=x+1 and x2+y2=13x^2+y^2=13? Trying individual points could miss a solution; substitution turns the two requirements into one equation in one unknown.

Simultaneous solution

A simultaneous solution is a set of values that satisfies every equation in a system at the same time. For two variables, write a solution as an ordered pair (x,y)(x,y).

A quadratic equation in two variables can contain terms such as x2x^2, xyxy and y2y^2, alongside linear and constant terms.

Use the linear equation to express one variable in terms of the other. Replace every occurrence of that variable in the quadratic equation. The result is usually a quadratic in the variable left. For example, if y=2x+1y=2x+1, then y2=(2x+1)2y^2=(2x+1)^2 and xy=x(2x+1)xy=x(2x+1).

Worked example 1: substituting a sum

Question. Show that the xx-coordinates of solutions to y=x+1y=x+1 and x2+y2=13x^2+y^2=13 satisfy x2+x−6=0x^2+x-6=0.

Plan. Replace yy by x+1x+1, expand the whole square and collect terms.

x2+(x+1)2=13substitute for yx2+x2+2x+1=13expand the square2x2+2x−12=0collect on one sidex2+x−6=0divide by 2\begin{aligned} x^2+(x+1)^2&=13 &&\text{substitute for }y\\ x^2+x^2+2x+1&=13 &&\text{expand the square}\\ 2x^2+2x-12&=0 &&\text{collect on one side}\\ x^2+x-6&=0 &&\text{divide by }2 \end{aligned}

Answer. Every simultaneous solution has an xx-coordinate satisfying x2+x−6=0x^2+x-6=0.

Check. At x=2x=2, the line gives y=3y=3. The original quadratic gives 22+32=132^2+3^2=13, and the reduced equation gives 4+2−6=04+2-6=0, so this candidate is preserved by the substitution.

Change the coefficient in the linear relation from 11 to 22, keeping the quadratic equation the same. The entire expression 2x+12x+1 must now be squared.

Worked example 2: a coefficient inside the square

Question. Show that the xx-coordinates of solutions to y=2x+1y=2x+1 and x2+y2=13x^2+y^2=13 satisfy 5x2+4x−12=05x^2+4x-12=0.

Plan. Substitute 2x+12x+1 for yy, then expand before collecting.

x2+(2x+1)2=13x2+4x2+4x+1=135x2+4x−12=0\begin{aligned} x^2+(2x+1)^2&=13\\ x^2+4x^2+4x+1&=13\\ 5x^2+4x-12&=0 \end{aligned}

Answer. The required equation is 5x2+4x−12=05x^2+4x-12=0.

Check. At x=−2x=-2, the line gives y=−3y=-3. Both original equations hold, and the reduced expression is 5(4)+4(−2)−12=05(4)+4(-2)-12=0. Substitution retains this candidate too.

The line can instead give xx in terms of yy. Choose the direction that keeps the algebra simple; a line such as x=3x=3 already gives a value to substitute.

Watch out

(2x+1)2=4x2+4x+1(2x+1)^2=4x^2+4x+1, not 4x2+14x^2+1. Missing the middle term changes the intersection points.

Reduce the next system without solving the quadratic yet.

Matching coordinate pairs

Solving the reduced quadratic finds one coordinate; substitute each root into the linear equation to recover its corresponding second coordinate.

Keep the roots separate as you substitute. Two possible xx-values and two possible yy-values do not give four solutions: each xx must stay with the yy produced by the same substitution. Check each completed pair in both original equations.

Worked example 3: recovering a y-coordinate for each root

Question. Solve the simultaneous equations y=x−1y=x-1 and x2+y2=25x^2+y^2=25.

Plan. Substitute for yy, solve the resulting quadratic in xx, then find the matching yy for each root.

x2+(x−1)2=252x2−2x−24=0x2−x−12=0(x−4)(x+3)=0\begin{aligned} x^2+(x-1)^2&=25\\ 2x^2-2x-24&=0\\ x^2-x-12&=0\\ (x-4)(x+3)&=0 \end{aligned}

Thus x=4x=4 or x=−3x=-3. Recover yy from the line:

x=4⇒y=4−1=3x=−3⇒y=−3−1=−4\begin{aligned} x=4&\quad\Rightarrow\quad y=4-1=3\\ x=-3&\quad\Rightarrow\quad y=-3-1=-4 \end{aligned}

Answer. (x,y)=(4,3)(x,y)=(4,3) or (−3,−4)(-3,-4).

Check. For (4,3)(4,3), 3=4−13=4-1 and 42+32=254^2+3^2=25. For (−3,−4)(-3,-4), −4=−3−1-4=-3-1 and (−3)2+(−4)2=25(-3)^2+(-4)^2=25. Both pairs satisfy both equations.

Now use the line to recover xx from each yy-root. The quadratic contains xyxy, so substitute into both the square and the product.

Worked example 4: recovering x instead

Question. Solve the simultaneous equations x=y+1x=y+1 and x2+xy=15x^2+xy=15.

Plan. Replace xx by y+1y+1, solve for yy, then recover each matching xx.

(y+1)2+(y+1)y=15substitute in both termsy2+2y+1+y2+y=152y2+3y−14=0(2y+7)(y−2)=0\begin{aligned} (y+1)^2+(y+1)y&=15 &&\text{substitute in both terms}\\ y^2+2y+1+y^2+y&=15\\ 2y^2+3y-14&=0\\ (2y+7)(y-2)&=0 \end{aligned}

Thus y=2y=2 or y=−7/2y=-7/2. Using x=y+1x=y+1,

y=2⇒x=3y=−72⇒x=−52\begin{aligned} y=2&\quad\Rightarrow\quad x=3\\ y=-\frac72&\quad\Rightarrow\quad x=-\frac52 \end{aligned}

Answer. (x,y)=(3,2)(x,y)=(3,2) or (−5/2,−7/2)(-5/2,-7/2).

Check. The first pair gives 3=2+13=2+1 and 32+3(2)=153^2+3(2)=15. The second gives −5/2=−7/2+1-5/2=-7/2+1 and

(−52)2+(−52)(−72)=254+354=15.\left(-\frac52\right)^2+\left(-\frac52\right)\left(-\frac72\right)=\frac{25}{4}+\frac{35}{4}=15.

Both satisfy the original system.

The reduced roots have to be translated back into the original variables before the simultaneous equation is fully solved.

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