6.03c - Simultaneous equations with quadratics

6.03c - Simultaneous equations with quadratics

A line and a curve can share two points, so simultaneous equations may have two different solution pairs. You will substitute to obtain a quadratic, keep each coordinate with its matching partner, and form equations for pairs of unknown numbers.

Substituting into a quadratic relation

Can y=x2y=x^2 and y=x+2y=x+2 both be true for two different pairs of numbers?

Higher only. A simultaneous solution must satisfy both equations. Since both expressions equal the same yy, they must equal each other: x2=x+2x^2=x+2. This is the same substitution idea used for linear pairs, but now the resulting equation is quadratic.

Each real xx-root can give a matching yy-value. On a coordinate diagram, a shared point lies on both graphs, so its coordinates satisfy both equations. We will find the pairs algebraically.

[DIAGRAM: asset_name: Two shared points of a line and a curve; asset_slug: line-quadratic-pairs; caption: Each labelled point satisfies both y = x squared and y = x + 2.; description: A coordinate graph shows the parabola y equals x squared and the line y equals x plus two. Their two intersections are labelled negative one, one and two, four. Pale dashed guides help match each horizontal x-coordinate with its own vertical y-coordinate.]
Each labelled point satisfies both y = x squared and y = x + 2.

Worked example 1: equating two expressions for y

Question. Solve y=x2y=x^2 and y=x+2y=x+2 simultaneously.

Plan. Equate the two expressions, solve for xx, then calculate yy for each root.

x2=x+2both expressions equal yx2−x−2=0rearrange to zero(x−2)(x+1)=0factorisex=2orx=−1solve factorsy=4ory=1use y=x+2 respectively\begin{aligned} x^2&=x+2 &&\text{both expressions equal }y\\ x^2-x-2&=0 &&\text{rearrange to zero}\\ (x-2)(x+1)&=0 &&\text{factorise}\\ x&=2\quad\text{or}\quad x=-1 &&\text{solve factors}\\ y&=4\quad\text{or}\quad y=1 &&\text{use }y=x+2\text{ respectively} \end{aligned}

Answer. (x,y)=(2,4)(x,y)=(2,4) or (−1,1)(-1,1).

Check. At (2,4)(2,4), 22=4=2+22^2=4=2+2. At (−1,1)(-1,1), (−1)2=1=−1+2(-1)^2=1=-1+2. Each complete pair satisfies both originals.

Change only the line's constant from two to six. Its slope stays the same, but the shared points move.

Worked example 2: changing the line's constant

Question. Solve y=x2y=x^2 and y=x+6y=x+6 simultaneously.

Plan. Equate, factorise, and substitute each xx-root separately.

x2=x+6equate the y-expressionsx2−x−6=0rearrange(x−3)(x+2)=0factorisex=3orx=−2find the rootsy=9ory=4use y=x+6 respectively\begin{aligned} x^2&=x+6 &&\text{equate the }y\text{-expressions}\\ x^2-x-6&=0 &&\text{rearrange}\\ (x-3)(x+2)&=0 &&\text{factorise}\\ x&=3\quad\text{or}\quad x=-2 &&\text{find the roots}\\ y&=9\quad\text{or}\quad y=4 &&\text{use }y=x+6\text{ respectively} \end{aligned}

Answer. (3,9)(3,9) or (−2,4)(-2,4).

Check. 32=9=3+63^2=9=3+6, and (−2)2=4=−2+6(-2)^2=4=-2+6. The new pairs also fit the same parabola.

If the quadratic has discriminant zero, the branches give one distinct pair. If it has negative discriminant, there is no real pair. For example, y=x2y=x^2 with y=−1y=-1 would require x2=−1x^2=-1, impossible for real xx. Finding two solutions is possible, not compulsory.

Watch out

Do not stop at two xx-roots. A simultaneous solution includes the corresponding yy, and both values must satisfy both equations.

Use the changed constant in the line to find both pairs.

Matching the two coordinates

Two xx-values and two yy-values do not create four freely interchangeable pairs. For each xx, substitution determines its own yy.

A small table is a safe way to preserve the match. Exact surd answers need the same care: the plus branch of an xx-root does not automatically mean the plus branch of yy. The substituted expression decides the signs.

Worked example 3: matching integer coordinates

Question. Solve y=x2y=x^2 and y=2x+3y=2x+3, presenting matched pairs.

Plan. Solve the quadratic, then make one row for each xx-value.

x2=2x+3equate expressions(x−3)(x+1)=0rearrange and factorisex=3orx=−1solve factors\begin{aligned} x^2&=2x+3 &&\text{equate expressions}\\ (x-3)(x+1)&=0 &&\text{rearrange and factorise}\\ x&=3\quad\text{or}\quad x=-1 &&\text{solve factors} \end{aligned}
xxy=2x+3y=2x+3Pair (x,y)(x,y)
3399(3,9)(3,9)
−1-111(−1,1)(-1,1)

Answer. (3,9)(3,9) or (−1,1)(-1,1).

Check. 32=93^2=9 and (−1)2=1(-1)^2=1, so both also satisfy the quadratic relation. The crossed pair (3,1)(3,1) fails both: 32≠13^2\ne1 and 2(3)+3≠12(3)+3\ne1.

Change only the line's constant from three to four. The roots become surds, but matching uses exactly the same substitution.

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