Solving a Quadratic Inequality

Critical values first, then pick the side.

Find the critical values first

Solve x² − 5x + 4 < 0. Solving an inequality means finding every value of x that makes it true. For a quadratic inequality that is usually a whole stretch of the number line, not one or two numbers.

Start by factoring: x² − 5x + 4 = (x − 1)(x − 4). The bracket x − 1 is 0 when x = 1, and x − 4 is 0 when x = 4. These two values are the critical values. They matter because the expression can only change from positive to negative, or back, by passing through 0, and it is 0 only at x = 1 and x = 4.

Sketch the curve

Now picture the graph of y = x² − 5x + 4. The coefficient of x² is positive, so the curve opens upward, like a bowl. It crosses the x-axis at the critical values, 1 and 4.

The inequality asks where x² − 5x + 4 is less than 0, which is where the curve is below the x-axis. A bowl dips below the axis only between its two crossings, so that is between 1 and 4. Test a value there to be sure: at x = 2, the value is 4 − 10 + 4 = −2, which is negative.

14y = (x − 1)(x − 4) < 0x = 6: y = 10 ✗y > 0y < 0turn it over

y < 0 holds between the roots, 1 < x < 4, because the curve is a bowl and that is where it is on that side of the axis: one interval, not two

Choose y < 0 and park x where it holds

The curve is y = (x − 1)(x − 4), which is x² − 5x + 4. With y < 0 chosen, the gold stretch of the axis is where the inequality holds. Drag the test point x into it and read the value of y there.

Write the answer

The expression is negative for every x between 1 and 4, so the solution is 1 < x < 4. The critical values themselves are not included: at x = 1 and x = 4 the expression is exactly 0, and 0 is not less than 0.

On a number line, the ends are open circles, because they are not part of the solution.

-101234561 < x < 4

The solution of x² − 5x + 4 < 0 is the stretch between the critical values, with both ends left out.

The other side of the axis

Reverse the sign: solve x² − 5x + 4 > 0. Now you want the curve above the x-axis. The bowl is above the axis to the left of 1 and to the right of 4. Test a value in each part: at x = 0 the value is 4, and at x = 5 it is 25 − 25 + 4 = 4. Both are positive.

So the solution is x < 1 or x > 4: two separate rays. They are joined by "or", because no number is both less than 1 and greater than 4, so "and" would describe no numbers at all.

-10123456x < 1 or x > 4

The solution of x² − 5x + 4 > 0 is the two rays outside the critical values.

Including the critical values

With ≤ or ≥, the critical values are included, because there the expression is 0, and 0 ≤ 0 is true. So x² − 5x + 4 ≤ 0 has the solution 1 ≤ x ≤ 4, and x² − 5x + 4 ≥ 0 has the solution x ≤ 1 or x ≥ 4. On a number line the ends become filled circles.

When the x² term is negative

Solve −x² + 5x − 4 > 0. The curve y = −x² + 5x − 4 opens downward, so it is above the axis between its crossings. One way to handle it is to multiply both sides by −1 first. Multiplying an inequality by a negative number reverses the sign, so the inequality becomes x² − 5x + 4 < 0, and the answer is 1 < x < 4, as before.

When it does not factor

The critical values are the roots of the quadratic, so when it does not factor, use the quadratic formula. For x² − 2x − 1 < 0, the roots are x = (2 ± √(4 + 4))/2 = 1 ± √2. The curve is a bowl, so the solution is 1 − √2 < x < 1 + √2, which is about −0.41 < x < 2.41.

The usual mistakes

Treating each bracket as in an equation: from (x − 1)(x − 4) < 0, writing x − 1 < 0 or x − 4 < 0. A product is negative when its two factors have opposite signs, not when either factor is negative. The critical values only mark where the sign can change; a sketch or a test value decides which side you want.

Taking the square root of both sides of x² < 9 to get x < 3. The number −5 is less than 3, but (−5)² = 25 is not less than 9. Factor instead: (x + 3)(x − 3) < 0, so −3 < x < 3.

Writing the two rays as 4 < x < 1. That says x is more than 4 and less than 1 at the same time, which no number is. Write x < 1 or x > 4.

Worked example: The Range of Prices for Which a Stall Makes a Profit

Question A drinks stall sells a cup of juice for x dollars. Its profit for a day is P dollars, where P = −5x2 + 60x − 100. (a) For which prices does the stall make a profit? (b) Which price gives the greatest profit, and how much is that profit?

  1. 1.The stall makes a profit when P > 0, so −5x2 + 60x − 100 > 0. Divide both sides by −5. Dividing by a negative number reverses the inequality sign: x2 − 12x + 20 < 0.

    −100−50050100024681012price in dollars, xprofit in dollars, Pa profit: −5x2+ 60x − 100 > 0divide both sides by −5 and reverse the sign:x2− 12x + 20 < 0
    −100−50050100024681012price in dollars, xprofit in dollars, Pa profit: −5x2+ 60x − 100 > 0divide both sides by −5 and reverse the sign:x2− 12x + 20 < 0
    The stall makes a profit when P > 0. Dividing both sides by −5 reverses the sign: x2 − 12x + 20 < 0.

  2. 2.Factorize: two numbers with a product of 20 and a sum of −12 are −2 and −10, so (x − 2)(x − 10) < 0. The profit is exactly zero at the roots x = 2 and x = 10.

    −100−50050100024681012price in dollars, xprofit in dollars, Px = 2x = 10(x − 2)(x − 10) < 0the profit is zero at x = 2 and at x = 10
    −100−50050100024681012price in dollars, xprofit in dollars, Px = 2x = 10(x − 2)(x − 10) < 0the profit is zero at x = 2 and at x = 10
    Factorize: (x − 2)(x − 10) < 0. The profit is exactly zero at x = 2 and at x = 10.

  3. 3.The coefficient of x2 in P is negative, so the graph of P opens downward, and it is above the x-axis between the roots. Test a price in each part: at x = 6, P = −180 + 360 − 100 = 80, and at x = 1 and at x = 11, P = −45.

    −100−50050100024681012price in dollars, xprofit in dollars, Px = 2x = 10P = 80P = −45P = −45the graph opens downward: above the axis between the rootsx = 6 gives P = 80; x = 1 and x = 11 give P = −45
    −100−50050100024681012price in dollars, xprofit in dollars, Px = 2x = 10P = 80P = −45P = −45the graph opens downward: above the axis between the rootsx = 6 gives P = 80; x = 1 and x = 11 give P = −45
    The graph of P opens downward, so it is above the x-axis between the roots: P = 80 at x = 6, and P = −45 at x = 1 and at x = 11.

  4. 4.(a) The stall makes a profit when 2 < x < 10, that is, when a cup costs more than $2 and less than $10.

    −100−50050100024681012price in dollars, xprofit in dollars, Px = 2x = 102 < x < 10a profit when 2 < x < 10more than $2 and less than $10 a cup
    −100−50050100024681012price in dollars, xprofit in dollars, Px = 2x = 102 < x < 10a profit when 2 < x < 10more than $2 and less than $10 a cup
    (a) The stall makes a profit when 2 < x < 10.

  5. 5.(b) The highest point of the graph is halfway between the roots, at x = 2 + 102 = 6, where P = 80. The greatest profit is $80 a day, at a price of $6 a cup.

    −100−50050100024681012price in dollars, xprofit in dollars, Px = 2x = 102 < x < 10(6, 80)the highest point is halfway between the roots: x = 6P = −180 + 360 − 100 = 80: a profit of $80
    −100−50050100024681012price in dollars, xprofit in dollars, Px = 2x = 102 < x < 10(6, 80)the highest point is halfway between the roots: x = 6P = −180 + 360 − 100 = 80: a profit of $80
    (b) The highest point is halfway between the roots, at x = 6, where the profit is $80.

Answer: (a) 2 < x < 10: a price of more than $2 and less than $10; (b) a price of $6, which gives a profit of $80

Common mistakes

  • Dividing by −5 and keeping the sign as >. Dividing both sides of an inequality by a negative number reverses the sign, so x2 − 12x + 20 must be less than zero.
  • Answering x < 2 or x > 10. That is where x2 − 12x + 20 is positive, which is where the profit is negative. A test price such as x = 6, where P = 80, shows which part of the number line is wanted.

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