A19 - Linear and quadratic simultaneous equations (Higher)

A19 - Linear and quadratic simultaneous equations (Higher)

When a line meets a quadratic curve, there can be two intersection points. You will use substitution to find both pairs of coordinates, including exact surd answers, and use graphs for approximate solutions. Each pair must satisfy both of the original equations.

Substituting a line into a quadratic (Higher)

The line y=x+1y=x+1 meets the curve y=x2−5y=x^2-5. How many points do they share, and what are their coordinates? Solving just the curve's equation would miss the point: we need values that lie on both.

You already know

A simultaneous solution is a pair that satisfies both equations. Substitute a complete expression for a variable, using brackets where needed. You can solve a quadratic by factorising, completing the square or the quadratic formula.

At an intersection, the two expressions for yy are equal. Set x+1=x2−5x+1=x^2-5, then solve the resulting equation in xx. This removes yy, just as substitution did with two linear equations. The difference is that the remaining equation is quadratic and may have two roots.

For every root, return to the linear equation to work out its matching yy. Two xx-values and two yy-values must be paired correctly. This substitution method is not on the formulae sheet; the quadratic formula is on the 2027 Higher formulae sheet.

Worked example 1: two integer intersection points

Question. Solve y=x+1y=x+1 and y=x2−5y=x^2-5. Non-calculator.

Plan. Equate the expressions for yy, factorise, then find the matching yy for each root.

x+1=x2−5same y at an intersectionx2−x−6=0collect on one side(x−3)(x+2)=0factorisex=3 or −2zero-product rulex=3:y=3+1=4first pairx=−2:y=−2+1=−1second pair\begin{aligned} x+1&=x^2-5&&\text{same }y\text{ at an intersection}\\ x^2-x-6&=0&&\text{collect on one side}\\ (x-3)(x+2)&=0&&\text{factorise}\\ x&=3\text{ or }-2&&\text{zero-product rule}\\ x=3:&\quad y=3+1=4&&\text{first pair}\\ x=-2:&\quad y=-2+1=-1&&\text{second pair} \end{aligned}

Why substitute each root separately? Each intersection has its own partner yy; mixing them creates points on neither graph.

Answer. (x,y)=(3,4)(x,y)=(3,4) or (−2,−1)(-2,-1).

Check. 4=3+1=32−54=3+1=3^2-5, and −1=−2+1=(−2)2−5-1=-2+1=(-2)^2-5.

Change only the line's constant term from +1+1 to −1-1. The curve stays fixed, but the intersections now have irrational coordinates.

Worked example 2: exact surd pairs

Question. Solve y=x−1y=x-1 and y=x2−5y=x^2-5. Give exact answers. Non-calculator.

Plan. Form the quadratic, use the formula, then subtract 11 from each xx.

x−1=x2−5equate the outputsx2−x−4=0rearrangex=1±(−1)2−4(1)(−4)2use the formula=1±172simplifyy=−1±172subtract 1\begin{aligned} x-1&=x^2-5&&\text{equate the outputs}\\ x^2-x-4&=0&&\text{rearrange}\\ x&=\frac{1\pm\sqrt{(-1)^2-4(1)(-4)}}2&&\text{use the formula}\\ &=\frac{1\pm\sqrt{17}}2&&\text{simplify}\\ y&=\frac{-1\pm\sqrt{17}}2&&\text{subtract }1 \end{aligned}

Why use matching signs? The plus value of xx gives the plus value of yy, because y=x−1y=x-1.

Answer. The pairs are

(1+172,−1+172)and(1−172,−1−172)\left(\frac{1+\sqrt{17}}2,\frac{-1+\sqrt{17}}2\right) \quad\text{and}\quad \left(\frac{1-\sqrt{17}}2,\frac{-1-\sqrt{17}}2\right)

Check. Each pair has y=x−1y=x-1. Each xx also obeys x2=x+4x^2=x+4, so x2−5=x−1=yx^2-5=x-1=y.

Two lines can share at most one point unless they coincide. A line and a quadratic curve can share two. The opening pair of graphs does exactly that: (3,4)(3,4) and (−2,−1)(-2,-1) both belong to both graphs.

Watch out

Wrong: solving x2−5=0x^2-5=0 to find the intersections with y=x+1y=x+1. That finds where the curve meets the horizontal axis. Right: set x2−5=x+1x^2-5=x+1, because the two graphs must have the same height.

Change the line's constant again and keep track of both pairs.

Substitution when both variables are squared (Higher)

The quadratic equation need not already be written as y=an expressiony=\text{an expression}. A linear equation still lets you replace one variable and create a quadratic in the other.

For x2+y2=25x^2+y^2=25 and y=x+1y=x+1, replace y2y^2 by (x+1)2(x+1)^2. The whole substituted expression must be squared. Use your bracket-expansion skills before collecting terms. The substitution and expansion rules are not on the formulae sheet: learn them.

Worked example 3: a square of a substituted expression

Question. Solve x2+y2=25x^2+y^2=25 and y=x+1y=x+1. Non-calculator.

Plan. Substitute for yy, expand its square, then solve and pair the values.

x2+(x+1)2=25substitute for yx2+x2+2x+1=25expand the square2x2+2x−24=0collect and rearrange(x+4)(x−3)=0halve then factorisex=−4 or 3solvey=−3 or 4use y=x+1\begin{aligned} x^2+(x+1)^2&=25&&\text{substitute for }y\\ x^2+x^2+2x+1&=25&&\text{expand the square}\\ 2x^2+2x-24&=0&&\text{collect and rearrange}\\ (x+4)(x-3)&=0&&\text{halve then factorise}\\ x&=-4\text{ or }3&&\text{solve}\\ y&=-3\text{ or }4&&\text{use }y=x+1 \end{aligned}

Why square a bracket? yy is the entire expression x+1x+1, so y2=(x+1)(x+1)y^2=(x+1)(x+1), not x2+1x^2+1.

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

Check. 16+9=2516+9=25 and −3=−4+1-3=-4+1; 9+16=259+16=25 and 4=3+14=3+1.

Change only the constant 2525 to 55. The same substitution now gives a smaller pair of intersection points.

Worked example 4: the same square, a new constant

Question. Solve x2+y2=5x^2+y^2=5 and y=x+1y=x+1. Non-calculator.

Plan. Use the same substitution, then factorise the new quadratic.

x2+(x+1)2=5substitute2x2+2x−4=0expand and rearrange(x+2)(x−1)=0halve then factorisex=−2 or 1solvey=−1 or 2find partners\begin{aligned} x^2+(x+1)^2&=5&&\text{substitute}\\ 2x^2+2x-4&=0&&\text{expand and rearrange}\\ (x+2)(x-1)&=0&&\text{halve then factorise}\\ x&=-2\text{ or }1&&\text{solve}\\ y&=-1\text{ or }2&&\text{find partners} \end{aligned}

Why check in both originals? The squared equation alone would also accept (2,1)(2,1), but that pair does not satisfy y=x+1y=x+1.

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

Check. 4+1=54+1=5, with −1=−2+1-1=-2+1; 1+4=51+4=5, with 2=1+12=1+1.

Continue this lesson.

You’re previewing NovaLearn without an account. Create a free account to read the rest.

Create free account