Tag Archives: quadratics

November 11, 2025 · 10:19 am

Completing the Square

x^2+6x-4 \rightarrow (x+3)^2-13

Completing the square is useful to

  • sketch parabolas.
  • solve quadratics.
  • factorising quadratics
  • finding the centre and radius version of the equation of a circle.

When completing the square we take advantage of perfect squares. For example, (x+3)^2=(x+3)(x+3)=x^2+6x+9

6=2\times 3 and 9=3\times 3

Example 1

Put x^2+8x-5 into completed square form.

What perfect square has an 8x term?

(x+4)^2=x^2+8x+16

We don’t want +16, we want -5, so subtract 16+5

x^2+8x-5=(x+4)^2-21

x^2+bx+c=(x+\frac{b}{2})^2-(\frac{b}{2})^2+c

What about a non-monic quadratic? For example,

2x^2+12x+11

Factorise the 2

2(x^2+6x+\frac{11}{2})

And continue as before

2[(x+3)^2-9+\frac{11}{2}]=2[(x+3)^2-\frac{18}{2}+\frac{11}{2}]=2[(x+3)^2-\frac{7}{2}]=2(x+3)^2-7

Example 2

y=2x^2+7x-5

2(x^2+\frac{7}{2}x-\frac{5}{2})

2[(x+\frac{7}{4})^2-(\frac{7}{4})^2-\frac{5}{2}]

2[(x+\frac{7}{4})^2-\frac{49}{16}-\frac{40}{16}]

2[(x+\frac{7}{4})^2-\frac{89}{16}]

2(x+\frac{7}{4})^2-\frac{89}{8}

ax^2+bx+c=a(x+\frac{b}{2a})^2-a((\frac{b}{2a})^2+\frac{c}{a})

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Filed under Algebra, Arithmetic, Classpad Skills, Completing the Square, Fractions, Quadratic, Quadratics, Year 10 Mathematics, Year 9 Mathematics

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May 9, 2024 · 6:08 am

Intersections of Lines and Circles

The three possibilities: no intersection, one point of intersection, or two points of intersection

Once my students have studied circle and quadratic equations, I like to introduce lines intersecting with circles. It tests their quadratic equation solving skills, and if they can use the discriminant.

Example 1

Calculate the point(s) of intersection of

    \[3x+2y-6=0\ \textnormal{and}\ (x+4)^2+(y+2)^2=9\]

Rearrange the line equation

    \[y=\frac{6-3x}{2}\]

    \[y=3-\frac{3x}{2}\]

Substitute for y into the circle equation

    \[(x+4)^2+(3-\frac{3x}{2}+2)^2=9\]

    \[(x+4)^2+(5-\frac{3x}{2})^2=9\]

    \[x^2+8x+16+25-15x+\frac{9x^2}{4}-9=0\]

    \[\frac{13x^2}{4}-7x+32=0\]

    \[13x^2-28x+128=0\]

At this point I like to check the discrimnant

    \[\Delta=b^2-4ac\]

    \[\Delta=(-28)^2-4\times13\times128\]

    \[\Delta=-5872\]

As the discriminate is less than zero, there are no points of intersection.

Example 2

Calculate the point(s) of intersection of

    \[y=2x+1\ \textnormal{and}\ (x+5)^2+(y+3)^2=16\]

    \[(x+5)^2+(2x+1+3)^2=16\]

    \[(x+5)^2+(2x+4)^2=16\]

    \[x^2+10x+25+4x^2+16x+16-16=0\]

    \[5x^2+26x+25=0\]

Check the discriminant

    \[\Delta=26^2-4\times5\times25\]

    \[\Delta=176\]

Therefore there are two points of intersection

    \[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]

    \[x=\frac{-26\pm\sqrt{176}}{10}\]

    \[x=\frac{-26\pm4\sqrt{11}}{10}\]

    \[x=\frac{-13\pm2\sqrt{11}}{5}\]

Substitute x into y

    \[y=2(\frac{-13\pm2\sqrt{11}}{5})+1\]

    \[\textnormal{The two points are}\ (\frac{-13+2\sqrt{11}}{5},\frac{-21+4\sqrt{11}}{5})\]

    \[\textnormal{and}\ (\frac{-13-2\sqrt{11}}{5},\frac{-21-4\sqrt{11}}{5})\]

Example 3

Find the value of k so that the line and the circle intersect at one point (i.e. the line is a tangent to the circle)

    \[-kx+y-5=0\ \textnormal{and}\ (x-2)^2+(y-3)^2=\frac{36}{5}\]

    \[y=kx+5\]

    \[(x-2)^2+(kx+5-3)^2=\frac{36}{5}\]

    \[(x-2)^2+(kx+2)^2=\frac{36}{5}\]

    \[x^2-4x+4+k^2x^2+4kx+4-\frac{36}{5}=0\]

    \[(1+k^2)x^2+(4k-4)x+\frac{4}{5}=0\]

Find the discriminant (remember for one solution we want the discriminant to be zero)

    \[\Delta=b^2-4ac\]

    \[\Delta=(4k-4)^2-4\times(1+k^2)\times\frac{4}{5}\]

    \[\Delta=16k^2-32k+16-\frac{16}{5}-\frac{16k^2}{5}\]

    \[0=16k^2-32k+16-\frac{16}{5}-\frac{16k^2}{5} \[0=80k^2-160k+80-16-16k^2\]

    \[0=64k^2-160k+64\]

    \[0=32(2k^2-5k+2)\]

    \[0=2k^2-5k+2\]

    \[0=2k^2-4k-k+2\]

    \[0=2k(k-2)-1(k-2)\]

    \[0=(2k-1)(k-2)\]

    \[k=\frac{1}{2}\ \textnormal{and}\ k=2\]

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