Quadratic simultaneous equations: worked examples and method

Swap y

Take the expression for y from the line and put it into the curve.

One quadratic

After substituting, rearrange until one side is zero.

Both roots

Factor or use the formula. Keep both roots unless a root fails the original story.

Back to y

Put each x into the line. The two points are not the two x-values alone.

A picture

The points you found should lie on the parabola and on the line.

Missing the curve

A negative discriminant means the line does not meet the curve.

Touching the curve

A line that touches the curve at exactly one point is a tangent to it. After substituting, the quadratic then has one repeated root, so there is exactly one solution.

The line goes into the curve

For y=x+2 and y=x^2, replace y in the second equation: x+2=x^2. Rearrange to x^2-x-2=0, which factorises as (x-2)(x+1)=0. So x=2 or x=-1. The linear equation then gives y=4 and y=1. The pairs are (2, 4) and (-1, 1). Check in y=x^2: 2^2=4 and (-1)^2=1. Both pairs work.

Always take the expression from the linear equation, because it gives one letter in a simple form. A circle such as x^2+y^2=25 cannot be written as a simple y= expression, so it is the line you rearrange. If the line is 2x+y=5, write y=5-2x first, even if that takes a line of work. The sign on 2x is the detail that later becomes the middle of the quadratic, so get it right before you square or substitute.

Method: 1. Rearrange the linear equation so one letter is the subject. 2. Substitute that expression into the quadratic. 3. You should be left with one equation in one letter. 4. Solve it, then find the other letter from the line.

Squaring a bracket

A circle x^2+y^2=25 with the line y=x+1 is the same method with one extra expansion. Replace y: x^2+(x+1)^2=25. The square is x^2+2x+1, not x^2+1. Then x^2+x^2+2x+1=25, so 2x^2+2x-24=0, and x^2+x-12=0. The factors are (x+4)(x-3). Then x=-4 gives y=-3, and x=3 gives y=4.

Check both pairs in the circle, which is the equation you substituted into, and glance at the line as well. (-4)^2+(-3)^2=16+9=25 and -4+1=-3. 3^2+4^2=25 and 3+1=4. The points (3, -3) and (-4, 4) would be the classic mix-up of the pairs. Each x must be substituted separately. Do not attach a y to an x by habit.

Method: 1. Expand (x+a)^2 as x^2+2ax+a^2. 2. Collect like terms and make one side zero. 3. Divide by a common factor if every term has one. 4. Factorise or use the formula, then pair each x with its own y.

A parabola and a line

For y=2x+1 and y=x^2-2, set 2x+1=x^2-2. Then 0=x^2-2x-3=(x-3)(x+1). So x=3 or x=-1. The line gives y=7 and y=-1. Check in the parabola: 3^2-2=7 and 1-2=-1. Two intersections. If the quadratic had come out with a negative discriminant, the line would miss the curve and the answer would be "no real solutions", which is allowed. Do not invent a point.

Keep the algebra in descending powers and move everything to one side before you factorise. A common slip is to write 2x+1=x^2-2 as 2x+1-2=x^2 and lose the constant. Subtract 2x+1 from both sides in one go: 0=x^2-2-2x-1=x^2-2x-3. Writing that subtraction out stops the constant from being left behind.

Method: 1. A line and y=x^2 usually meet at two points, one point, or not at all. 2. The quadratic after substitution tells you which. 3. A repeated root is a tangent. 4. No real root means the line misses the curve.

How to present the pairs

The final answer is a list of coordinate pairs, not a list of x values. If the question says "find the points of intersection", write (2, 4) and (-1, 1). If it says "solve the simultaneous equations", the same pairs are the solution. Label them so a reader can see which y belongs to which x. A small table with a column for x and a column for y is clear.

On a non-calculator paper the quadratic at the end should factorise, or it should be a difference of squares, or the formula should stay exact. If you reach x^2+x-12, look for the pair 4 and -3 before you reach for a formula. After the pairs are written, one substitution into the untouched equation is enough to show the method was honest. Two substitutions are better when the numbers are small.

Method: 1. Write coordinates, not a list of x values alone. 2. Each x is paired with the y from the line. 3. Check both points in the curve as well. 4. Two roots mean two points. Do not mix the y values between them.

Rearranging the linear equation before you substitute

The linear equation is not always given as y=ldots. You may see x+y=8 or 2x+y=7. Rearrange it so one letter is on its own, for example y=8-x, before you substitute into the quadratic or circle. Choose the letter that is easiest to isolate, and avoid fractions where you can. This step is routine, yet a slip here carries through the rest of the question, so write it clearly.

Put the whole expression into the other equation in brackets. If the second equation is y=x^2-3x, the substitution gives 8-x=x^2-3x. Moving everything to one side gives x^2-2x-8=0. Brackets matter most when you substitute into a circle, because y^2 with y=x-2 becomes (x-2)^2, not x^2-2^2. Slow, tidy algebra at this stage earns the method marks even when the final answer slips.

Method: 1. Rearrange the linear equation so y (or x) is the subject. 2. Write the substitution with brackets around the whole expression. 3. Expand brackets and collect every term on one side so the other side is 0. 4. Divide through by any common factor to make the quadratic simpler.

Tip: Dividing a quadratic through by a common factor does not change its roots and makes factorising easier.

Show that, and writing the answer as coordinate pairs

Many exam questions stage the work. Part (a) says show that the x values satisfy a given quadratic, and part (b) says hence find the coordinates. Part (a) is a substitution and expansion with the target written out, so every line has to be clear. Part (b) uses that quadratic, so do not start again.

Once you have the x values, substitute each into the linear equation, which is the simpler one, to find the matching y value. Pair them up carefully. Writing them as points (5, 1) and (-1, -5) shows the pairing. Check that each pair satisfies both original equations, because it is easy to pair the wrong y with an x.

Method: 1. In a show that part, substitute, expand, and arrange into exactly the given quadratic. 2. Solve the quadratic to find the x values. 3. Find each y value using the linear equation. 4. Write the solutions as pairs and check each pair in the other equation.

Tangents, and how many solutions there are

After substituting you have a quadratic in x. If it has two different roots, the line cuts the curve at two points. If the quadratic has one repeated root, the line touches the curve at one point and is a tangent. If it has no real roots, the line misses the curve. The expression b^2-4ac tells you which case you have: positive for two points, zero for one, negative for none.

To show a line is a tangent, substitute, rearrange, and show the quadratic factorises as a perfect square such as (x+4)^2=0 or that b^2-4ac=0. Some questions include an unknown number, for example a line y=4x+c that touches a curve. Set b^2-4ac=0 and solve for c. Then find the point of contact from the repeated root.

Method: 1. Substitute and write the quadratic as ax^2+bx+c=0. 2. Work out b^2-4ac, or check for a perfect square. 3. Zero means a tangent, positive means two intersections, negative means none. 4. For an unknown constant, put b^2-4ac=0 and solve.

Lengths between the points, and worded problems

Once you have the two intersection points A and B, a question may ask for the length AB. Use Pythagoras on the difference in the x values and the difference in the y values. For A(5, 1) and B(-1, -5) both differences are 6, so AB=√(6^2+6^2)=√(72)=6√(2). Leave the answer in surd form if the question says so.

Worded problems also lead to a quadratic and a linear equation. A rectangle with perimeter 34 cm and area 60 cm^2 gives x+y=17 and xy=60. Substitute y=17-x to get x^2-17x+60=0, which factorises to give sides of 5 and 12. Check that the answers make sense in the context, and reject any negative length.

Method: 1. Write the linear equation from the words, such as half the perimeter. 2. Write the second equation from the area or product. 3. Substitute, solve the quadratic, and find the matching values. 4. Reject answers that cannot be lengths, and check both conditions.