6.04a - Quadratic inequalities

6.04a - Quadratic inequalities

A quadratic can be positive for two separate ranges of inputs. Find its zero boundaries, determine its sign between and beyond them, then state the complete solution with the correct endpoints and set notation.

Higher only.

Finding the boundary roots

How can both −5-5 and 5 satisfy x2>9x^2>9, while zero does not? Squaring measures size without retaining the original sign, so “greater than” need not mean only inputs on the right.

For x2>9x^2>9, the expression being compared with zero is x2−9x^2-9. It is zero at −3-3 and 3. These are the boundary roots: the values where the expression may change sign. Finding them is the first step, not the complete inequality solution.

You already know

To find a quadratic's roots, make one side zero and choose factorisation, completing the square or the quadratic formula. An inequality also needs a decision about the intervals between the roots.

First rearrange the inequality so you can identify the whole quadratic whose sign matters.

Worked example 1: a quadratic already compared with zero

Question. Find the boundary roots for x2−x−6<0x^2-x-6<0.

Plan. Replace the comparison with equality and factorise.

x2−x−6=0find zero boundaries(x−3)(x+2)=0factorisex=3 or x=−2one factor is zero\begin{aligned} x^2-x-6&=0 &&\text{find zero boundaries} \\ (x-3)(x+2)&=0 &&\text{factorise} \\ x&=3\text{ or }x=-2 &&\text{one factor is zero} \end{aligned}

Answer. The boundary roots, in increasing order, are −2-2 and 3. They are excluded from the original strict inequality because the quadratic equals zero there.

Check. (−2)2−(−2)−6=0(-2)^2-(-2)-6=0 and 32−3−6=03^2-3-6=0. Also, expanding the factors returns x2−x−6x^2-x-6.

Change only the right-hand side. Rearrangement changes the constant term and therefore the roots.

Worked example 2: rearrange before finding roots

Question. Find the boundary roots for x2−x−6<6x^2-x-6<6.

Plan. Subtract six from both sides before solving the boundary equation.

x2−x−12<0subtract 6x2−x−12=0find zero boundaries(x−4)(x+3)=0factorisex=−3 or x=4solve each factor\begin{aligned} x^2-x-12&<0 &&\text{subtract 6} \\ x^2-x-12&=0 &&\text{find zero boundaries} \\ (x-4)(x+3)&=0 &&\text{factorise} \\ x&=-3\text{ or }x=4 &&\text{solve each factor} \end{aligned}

Answer. The boundary roots are −3-3 and 4, both excluded from the strict inequality.

Check. In the original expression, (−3)2−(−3)−6=6(-3)^2-(-3)-6=6 and 42−4−6=64^2-4-6=6. Each makes the two sides equal, exactly as a boundary should.

The first example has the factorisation (x−3)(x+2)(x-3)(x+2); the second has (x−4)(x+3)(x-4)(x+3). Using the first pair of roots after changing the right side would solve a different problem.

Watch out

Do not replace (x−3)(x+2)<0(x-3)(x+2)<0 with “x<3x<3 or x<−2x<-2”. A product is negative when its factors have opposite signs. Find the roots first; examine the product on each interval next.

For this check, stop at the boundaries rather than guessing the solution interval.

Testing signs between roots

Between two neighbouring roots, a quadratic has no zero at which to change sign. Test one input in each interval to establish that interval's sign.

For (x+2)(x−3)(x+2)(x-3), the factors have the same sign outside the roots and opposite signs between them. The sign pattern is therefore positive, negative, positive. A quick curved sketch can show the same idea: the height represents the expression's value, so above the horizontal axis means positive and below means negative. You do not need to plot a full graph to use this sign information.

Worked example 3: selecting the middle interval

Question. Solve (x+2)(x−3)<0(x+2)(x-3)<0.

Plan. Split the number line at −2-2 and 3, then test the product in each interval.

Intervalx<−2−2<x<3x>3Test input−304Product(−1)(−6)=6(2)(−3)=−6(6)(1)=6Sign+−+\begin{array}{c|c|c|c} \text{Interval}&x<-2&-2<x<3&x>3\\\hline \text{Test input}&-3&0&4\\ \text{Product}&(-1)(-6)=6&(2)(-3)=-6&(6)(1)=6\\ \text{Sign}&+&-&+ \end{array}

Answer. −2<x<3-2<x<3. The product is negative only between the two roots; zero at either root is not allowed.

Check. x=1x=1 gives 3×(−2)=−6<03\times(-2)=-6<0. Inputs −4-4 and 5 give positive products and are correctly excluded. The factor signs establish the whole intervals.

Change only the direction of the comparison. We keep the same boundaries and sign table but select the other intervals.

Worked example 4: selecting the outside intervals

Question. Solve (x+2)(x−3)>0(x+2)(x-3)>0.

Plan. Use the positive intervals from the same product sign pattern.

x<−2:(−)×(−)=+−2<x<3:(+)×(−)=−x>3:(+)×(+)=+\begin{aligned} x<-2&:\quad(-)\times(-)=+ \\ -2<x<3&:\quad(+)\times(-)=- \\ x>3&:\quad(+)\times(+)=+ \end{aligned}

Answer. x<−2x<-2 or x>3x>3. Both strict endpoints are excluded.

Check. x=−3x=-3 and x=4x=4 each give 6, a positive value. Zero gives −6-6, so the answer must not include the middle interval.

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