Technique

Common root of two quadratics by elimination

Algebra • Quadratic Equations • Roots

Statement

If a value of

x

satisfies both

a_{1} x^{2} + b_{1} x + c_{1} = 0

and

a_{2} x^{2} + b_{2} x + c_{2} = 0

, it satisfies every combination of the two; scaling so the

x^{2}

coefficients agree and subtracting leaves a linear equation, and its solution is the only value that can be the common root.

How to use it

1Match the $x^{2}$ coefficients and subtract: what survives is linear in $x$ .
2Substitute the resulting $x$ back into either quadratic to pin down an unknown parameter.
3Elimination gives a necessary condition only, so check the value it produces in the original equations.

Worked example

For which values of

k

x^{2} - 4 x + k = 0

and

x^{2} - 6 x + 3 k = 0

have a common root?

Step-by-step

1
$(x^{2} - 4 x + k) - (x^{2} - 6 x + 3 k) = 2 x - 2 k = 0 \Rightarrow x = k$
Why: Subtracting removes $x^{2}$ and leaves one linear equation.
2
$k^{2} - 4 k + k = 0 \Rightarrow k (k - 3) = 0$
Why: Substitute $x = k$ into the first quadratic.
3
$k = 0$ or $k = 3$ , and both check out in the second quadratic
Why: The candidates must be verified, not just derived.

Answer

k = 0

k = 3

Watch out

The linear equation says what the common root must be if one exists; only substituting back confirms that it does. If the two quadratics are proportional they share both roots and the subtraction collapses to

0 = 0

, telling you nothing.

Builds on

Quadratic formula

Also known as

common root of two quadratics, subtract two quadratics, shared factor of two quadratics, eliminate the x squared term

Common root of two quadratics by elimination

Statement

Worked example

Appears in past papers

2017 (II)