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Polynomials & quadratic roots · Lesson 3 of 3

Completing the square, the discriminant and inequalities

Write ax² + bx + c as p(x + q)² + r, solve quadratic inequalities with any sign in front, and find the range of k that gives real roots.

20 minYou should already know: Quadratics & their graphs Polynomials & algebraic division Inequalities
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This lesson joins three things you met in General Maths: completing the square and the discriminant (see quadratic equations), and quadratic inequalities (see inequalities). The Further Maths question that needs all of them is: for what values of kk does this equation have real roots? The answer is a quadratic inequality in kk.

Completing the square: the form p(x + q)² + r

Any quadratic can be written as p(x+q)2+rp(x + q)^2 + r. This form shows the turning point: the squared bracket is never negative, so for p>0p > 0 the lowest value is rr, when x=−qx = -q.

x(−q, r)x = −q
y = p(x + q)² + rLowest point (−q, r) when p > 0

When there is a number in front of x2x^2, take it out of the xx terms only. Don’t divide the whole expression by it: that changes its value.

Worked example · WAEC 2019

WAEC 2019 · Paper 2 · Q9 (a)

If y=2x2−5x+3y = 2x^2 - 5x + 3 is expressed in the form y=p(x+q)2+ry = p(x + q)^2 + r, where pp, qq and rr are constants, find (q+r)(q + r).

  1. Take out the 2

    • y=2(x2−52x)+3{y = 2\left(x^2 - \frac52x\right) + 3}.

    Think first. Take 2 out of 2x² − 5x only. What is left in the bracket?

  2. Complete the square inside

    • Half of −52-\frac52 is −54-\frac54.
    • x2−52x=(x−54)2−2516{x^2 - \frac52x = \left(x - \frac54\right)^2 - \frac{25}{16}}.

    Think first. Half of −5/2 is?

  3. Multiply back by 2

    • y=2[(x−54)2−2516]+3{y = 2\left[\left(x - \frac54\right)^2 - \frac{25}{16}\right] + 3}.
    • =2(x−54)2−258+3{= 2\left(x - \frac54\right)^2 - \frac{25}{8} + 3}.
    • −258+248=−18{-\frac{25}{8} + \frac{24}{8} = -\frac18}.
    • So y=2(x−54)2−18{y = 2\left(x - \frac54\right)^2 - \frac18}.

  4. Read off and add

    • p=2{p = 2}, q=−54{q = -\frac54}, r=−18{r = -\frac18}.
    • q+r=−108−18=−118{q + r = -\frac{10}{8} - \frac18 = -\frac{11}{8}}.

    Think first. Compare with p(x + q)² + r.

More: completing the square

The discriminant and real roots

For ax2+bx+c=0ax^2 + bx + c = 0, the discriminant b2−4acb^2 - 4ac tells you how many roots there are. Further Maths questions use exact words for each case:

b² − 4ac > 0two rootsb² − 4ac = 0one repeatedb² − 4ac < 0no real rootsreal roots: b² − 4ac ≥ 0
What b² − 4ac tells youReal roots means two different or equal: b² − 4ac ≥ 0
The question saysCondition
real rootsb2−4ac≥0b^2 - 4ac \ge 0
real and different (distinct) rootsb2−4ac>0b^2 - 4ac > 0
equal rootsb2−4ac=0b^2 - 4ac = 0
no real rootsb2−4ac<0b^2 - 4ac < 0

When a coefficient contains kk, the condition becomes an inequality in kk, and very often a quadratic inequality. So first, a closer look at those.

Quadratic inequalities

The General Maths method still works. Get 0 on one side, find the roots, then picture the curve:

xab< 0 between> 0> 0
Between or outsideFor a positive x² term: < 0 between the roots; > 0 outside them

If the x2x^2 term is negative, multiply through by −1-1 first and turn the inequality sign round. Then the picture above applies.

Worked example · WAEC 2019

WAEC 2019 · Paper 2 · Q3 (b)

Find the range of values of nn for which 3+14n−5n2≤03 + 14n - 5n^2 \le 0.

  1. Make the n² term positive

    • Write it in order: −5n2+14n+3≤0{-5n^2 + 14n + 3 \le 0}.
    • Multiply by −1-1 and turn the sign round: 5n2−14n−3≥0{5n^2 - 14n - 3 \ge 0}.

    Think first. Multiply by −1. What happens to ≤?

  2. Factorise

    • The numbers are −15-15 and 11.
    • Split the middle term: 5n2−15n+n−3{5n^2 - 15n + n - 3}.
    • Take out common factors in pairs: 5n(n−3)+1(n−3){5n(n - 3) + 1(n - 3)}.
    • So (5n+1)(n−3)≥0{(5n + 1)(n - 3) \ge 0}.

    Think first. Two numbers that multiply to 5 × (−3) = −15 and add to −14?

  3. Between or outside?

    • The roots are n=−15{n = -\frac15} and n=3{n = 3}.
    • "≥0\ge 0" is outside the roots: n≤−15{n \le -\frac15} or n≥3{n \ge 3}.

    Think first. The roots are −1/5 and 3. Is ≥ 0 between them or outside?

  4. Test a value

    • Try n=0n = 0, which is between the roots: 3+0−0=3{3 + 0 - 0 = 3}.
    • 3 is not ≤0\le 0, so the values between the roots are rightly left out.

More: inequalities

Real roots: finding the range of k

Now put the two together:

  1. rearrange the equation to ax2+bx+c=0ax^2 + bx + c = 0, and write down aa, bb and cc in terms of kk;
  2. write the condition, for example b2−4ac≥0b^2 - 4ac \ge 0 for real roots;
  3. multiply out and simplify, then solve the inequality in kk.

Slide kk and watch both graphs. The curve meets the xx-axis exactly when the discriminant is on or above the kk-axis:

Real roots: an inequality in kPick a family, slide k

y = x² + 2x + 9

−7−5−3−11357−8−44812xy

b² − 4ac = k² − 36

−10−8−6−4−2246810−40−20204060k−66
x² + 2x + 9 = 0the equation−32b² − 4acno real rootsso
For x² + kx + 9 = 0, b² − 4ac = k² − 36. Real roots need (k − 6)(k + 6) ≥ 0, so k ≤ −6 or k ≥ 6. This k is outside the range: the curve misses the x-axis, and the point on the lower graph is below its axis.

Worked example · WAEC 2014

WAEC 2014 · Paper 2 · Q2

If 2x2+3x+3=kx−k2x^2 + 3x + 3 = kx - k has real roots, find the range of values of kk.

  1. Rearrange

    • Take kxkx and −k-k to the left: 2x2+3x−kx+3+k=0{2x^2 + 3x - kx + 3 + k = 0}.
    • Group the xx terms: 2x2+(3−k)x+(3+k)=0{2x^2 + (3 - k)x + (3 + k) = 0}.
    • So a=2{a = 2}, b=3−k{b = 3 - k}, c=3+k{c = 3 + k}.

    Think first. Take kx − k to the left. What are a, b and c?

  2. The condition

    • Real roots: b2−4ac≥0{b^2 - 4ac \ge 0}.
    • Substitute: (3−k)2−4(2)(3+k)≥0{(3 - k)^2 - 4(2)(3 + k) \ge 0}.

    Think first. Real roots need b² − 4ac to be what?

  3. Simplify

    • Multiply out the square: (3−k)2=9−6k+k2{(3 - k)^2 = 9 - 6k + k^2}.
    • Multiply out the second part: 8(3+k)=24+8k{8(3 + k) = 24 + 8k}.
    • So 9−6k+k2−24−8k≥0{9 - 6k + k^2 - 24 - 8k \ge 0}.
    • Collect terms: k2−14k−15≥0{k^2 - 14k - 15 \ge 0}.

  4. Solve the inequality

    • Factorise: (k−15)(k+1)≥0{(k - 15)(k + 1) \ge 0}.
    • The roots are k=−1k = -1 and k=15k = 15, and "≥0\ge 0" is outside them.
    • So k≤−1{k \le -1} or k≥15{k \ge 15}.

    Think first. Factorise, then decide: between or outside?

When kk is in both aa and cc, the k2k^2 terms often cancel, and the inequality is linear:

More: real roots

Questions that end in a quadratic

Many Further Maths questions in other topics turn into a quadratic equation part way through: a determinant, a definite integral, a surd equation or a hidden power. Solve it as usual, then check which roots fit the question. A limit of an integral, a power such as 2y2^y, or a root of a surd equation may rule one out.

More: quadratics inside other questions

Your turn

WAEC 2019 · Paper 2 · Q10 (a)

  1. (a)

    Find the range of values of pp for which 4x2−px+1=04x^2 - px + 1 = 0 has real roots.

    Show the answer

    p≤−4p \le -4 or p≥4p \ge 4

Worked solution (try it first)

(a)

  1. Read off the coefficients: a=4a = 4, b=−pb = -p, c=1c = 1.
  2. Real roots need b2−4ac≥0b^2 - 4ac \ge 0: p2−16≥0p^2 - 16 \ge 0.
  3. Factorise: (p−4)(p+4)≥0(p - 4)(p + 4) \ge 0, with roots p=−4p = -4 and p=4p = 4.
  4. "≥0\ge 0" is outside the roots: p≤−4p \le -4 or p≥4p \ge 4.

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