Solving Quadratic Equations by Using the Quadratic Formula

Using the quadratic formula with \( a = 2 \), \( b = 3 \), \( c = -5 \): the discriminant is \( b^2 \; – \; 4ac = 9 \; – \; 4(2)(-5) = 9 + 40 = 49 \). Since the discriminant is positive, the ± in the formula produces two different values: \( x = \frac{-3 + 7}{4} = 1 \) and \( x = \frac{-3 \; – \; 7}{4} = \frac{-10}{4} = -2.5 \). We can verify: \( 2(1)^2 + 3(1) \; – \; 5 = 0 \) ✓ and \( 2(-2.5)^2 + 3(-2.5) \; – \; 5 = 12.5 \; – \; 7.5 \; – \; 5 = 0 \) ✓.

The ± symbol in the formula is what generates two solutions. A common mistake is the “only using one branch of ±” error — computing only \( \frac{-b + \sqrt{\text{disc}}}{2a} \) or only \( \frac{-b \; – \; \sqrt{\text{disc}}}{2a} \), giving just one solution. Every quadratic equation has the potential for two solutions — the discriminant determines whether those solutions are distinct, repeated, or non-real.

Using the formula with \( a = 1 \), \( b = -8 \), \( c = 16 \): the discriminant is \( (-8)^2 \; – \; 4(1)(16) = 64 \; – \; 64 = 0 \). So \( x = \frac{8 \pm \sqrt{0}}{2} = \frac{8}{2} = 4 \). Since \( \sqrt{0} = 0 \), the + and − paths give the same value. We can verify: \( 4^2 \; – \; 8(4) + 16 = 16 \; – \; 32 + 16 = 0 \) ✓. Notice that \( x^2 \; – \; 8x + 16 = (x \; – \; 4)^2 \), confirming this is a perfect square with a single repeated root.

Students often assume that because the formula contains ±, it must always produce two different answers. When the discriminant equals zero, the ± adds and subtracts nothing, so both routes collapse to a single solution. This is called a repeated root.

The quadratic formula requires the equation in the form \( ax^2 + bx + c = 0 \). Crucially, the Right Hand Side must be zero. In \( x^2 + 3 = 4x \), if a student tries to read off values directly, they might set \( a = 1 \), \( b = 3 \), \( c = 4 \) — but \( b \) is the coefficient of the \( x \) term, and there is no \( x \) term on the left as written. Rearranging gives \( x^2 \; – \; 4x + 3 = 0 \), so \( a = 1 \), \( b = -4 \), \( c = 3 \). The correct solutions are \( x = \frac{4 \pm \sqrt{16 \; – \; 12}}{2} = \frac{4 \pm 2}{2} \), giving \( x = 3 \) or \( x = 1 \).

Using the wrong values (e.g. \( b = 3 \)) would give a completely different discriminant and incorrect answers. The formula is derived from completing the square on \( ax^2 + bx + c = 0 \), so the equation must be in this form for \( a \), \( b \), and \( c \) to have their proper meaning.

With \( a = 1 \), \( b = -2 \), \( c = -5 \): the discriminant is \( 4 \; – \; 4(1)(-5) = 4 + 20 = 24 \). Since 24 is not a perfect square (\( 4^2 = 16 \), \( 5^2 = 25 \)), \( \sqrt{24} \) is irrational. The solutions are \( x = \frac{2 \pm \sqrt{24}}{2} = \frac{2 \pm 2\sqrt{6}}{2} = 1 \pm \sqrt{6} \). Since \( \sqrt{6} \approx 2.449 \), the two solutions are approximately 3.449 and −1.449 — neither is a whole number.

Note on precision: Students often rush to the calculator to get the decimal approximation. However, the form \( 1 \pm \sqrt{6} \) is the exact answer. It is better to keep the surd form until the very last step to avoid rounding errors.

Example: \( x^2 \; – \; 5x + 6 = 0 \) → \( x = \frac{5 \pm 1}{2} \), so \( x = 3 \) or \( x = 2 \).

Another: \( x^2 \; – \; 7x + 10 = 0 \) → \( x = \frac{7 \pm 3}{2} \), so \( x = 5 \) or \( x = 2 \).

Creative: \( x^2 \; – \; 3x + 1 = 0 \) → discriminant = 5, so \( x = \frac{3 \pm \sqrt{5}}{2} \approx 2.618 \) or \( \approx 0.382 \). Both positive, but neither is an integer — showing that “two positive solutions” doesn’t require nice numbers.

Trap: \( x^2 + 5x + 6 = 0 \) → \( x = \frac{-5 \pm 1}{2} \), so \( x = -2 \) or \( x = -3 \). Both solutions are negative. A student might see positive coefficients 5 and 6 and assume the solutions are positive, but the \( -b \) in the formula means the sign of \( b \) directly affects the sign of the solutions. To get positive solutions when \( a = 1 \), you need \( b < 0 \).

Example: \( x^2 \; – \; 4x + 4 = 0 \) → discriminant \( = 16 \; – \; 16 = 0 \), so \( x = 2 \) (repeated root).

Another: \( x^2 + 6x + 9 = 0 \) → discriminant \( = 36 \; – \; 36 = 0 \), so \( x = -3 \) (repeated root).

Creative: \( 4x^2 \; – \; 4x + 1 = 0 \) → discriminant \( = 16 \; – \; 16 = 0 \), so \( x = \frac{1}{2} \) (repeated root). This shows the discriminant can be zero even when \( a \neq 1 \) and the repeated root can be a fraction.

Trap: \( x^2 \; – \; 8x + 8 = 0 \) → discriminant \( = 64 \; – \; 32 = 32 \neq 0 \). A student might see the two 8s and assume they “cancel” in the discriminant, but \( b^2 \; – \; 4ac = (-8)^2 \; – \; 4(1)(8) = 64 \; – \; 32 \), not \( 64 \; – \; 64 \). For \( a = 1 \), the discriminant is zero only when \( c = \frac{b^2}{4} \) — here that would require \( c = 16 \), not 8.

Example: \( x^2 = 3x \; – \; 2 \) → rearranges to \( x^2 \; – \; 3x + 2 = 0 \), so \( a = 1 \), \( b = -3 \), \( c = 2 \).

Another: \( 2x^2 + 1 = 5x \) → rearranges to \( 2x^2 \; – \; 5x + 1 = 0 \), so \( a = 2 \), \( b = -5 \), \( c = 1 \).

Creative: \( x(x + 3) = 4 \) → expand first: \( x^2 + 3x = 4 \), then rearrange to \( x^2 + 3x \; – \; 4 = 0 \). This requires two steps (expand, then rearrange) before the formula can be used.

Trap: \( 3x^2 \; – \; 6x + 3 = 0 \). This is already in standard form (\( a = 3 \), \( b = -6 \), \( c = 3 \)), so no rearranging is needed. A student might think the 3 out front means it needs manipulation, but the formula can be applied directly.

Example: \( x^2 \; – \; 4x + 1 = 0 \) → discriminant \( = 12 \), so \( x = \frac{4 \pm 2\sqrt{3}}{2} = 2 \pm \sqrt{3} \).

Another: \( x^2 + 6x + 4 = 0 \) → discriminant \( = 20 \), so \( x = \frac{-6 \pm 2\sqrt{5}}{2} = -3 \pm \sqrt{5} \).

Creative: \( x^2 + 14x + 2 = 0 \) → discriminant \( = 196 \; – \; 8 = 188 = 4 \times 47 \), so \( x = \frac{-14 \pm 2\sqrt{47}}{2} = -7 \pm \sqrt{47} \). Note: The question asks for the form \( p \pm \sqrt{r} \). You can only reach this form if the denominator cancels out completely with a factor from the numerator (i.e. if \( b \) is even and the discriminant is a multiple of 4).

Trap: \( x^2 \; – \; 2x \; – \; 3 = 0 \) → discriminant \( = 4 + 12 = 16 = 4^2 \). Since the discriminant is a perfect square, the solutions are \( x = \frac{2 \pm 4}{2} = 3 \) or \( -1 \) — rational numbers, not surds. A student might see the −3 and think the formula will give something messy, but the discriminant being a perfect square means the solutions simplify neatly.

This depends entirely on the discriminant \( b^2 \; – \; 4ac \). When the discriminant is positive, the equation has two distinct real solutions — for example, \( x^2 + x \; – \; 6 = 0 \) has discriminant 25, giving \( x = 2 \) and \( x = -3 \). When the discriminant is zero, there is exactly one repeated solution — for example, \( x^2 \; – \; 8x + 16 = 0 \) gives only \( x = 4 \). When the discriminant is negative, there are no real solutions at all — for example, \( x^2 + 1 = 0 \) has discriminant \( -4 \).

Students often believe the quadratic formula “always gives two answers” because they see the ± symbol. The visual connection is key:

When \( c = 0 \), the equation becomes \( ax^2 + bx = 0 \), which factors as \( x(ax + b) = 0 \). This immediately gives \( x = 0 \) or \( x = -\frac{b}{a} \) — no formula needed. For example, \( 3x^2 + 12x = 0 \) factors as \( 3x(x + 4) = 0 \), giving \( x = 0 \) or \( x = -4 \). This factorisation works for any values of \( a \) and \( b \) (provided \( a \neq 0 \)).

Students sometimes reach for the quadratic formula automatically without checking whether simpler methods work. Spotting that \( c = 0 \) means every term contains \( x \), so \( x \) factors out immediately. The formula will give the same answers, but factorising is faster and less error-prone.

The formula gives integer solutions when two conditions are met: the discriminant \( b^2 \; – \; 4ac \) must be a perfect square, and \( -b \pm \sqrt{b^2 \; – \; 4ac} \) must be divisible by \( 2a \). For example, \( x^2 \; – \; 7x + 12 = 0 \) has discriminant \( 49 \; – \; 48 = 1 \) and gives \( x = \frac{7 \pm 1}{2} \), so \( x = 4 \) and \( x = 3 \) (integers). But \( x^2 \; – \; 6x + 2 = 0 \) has discriminant \( 36 \; – \; 8 = 28 \), so the solutions are \( 3 \pm \sqrt{7} \) (irrational, not integers).

Students often expect “nice” answers from the quadratic formula and become unsettled when surds appear. In fact, most randomly chosen quadratic equations produce irrational solutions — integer solutions are the special case, not the norm.

A negative discriminant means there are no real solutions — not one. The formula gives \( x = \frac{-b \pm \sqrt{\text{negative}}}{2a} \), and the square root of a negative number is not a real number. For example, \( x^2 + 2x + 5 = 0 \) has discriminant \( 4 \; – \; 20 = -16 \), and there is no real value of \( x \) that satisfies the equation. The graph of \( y = x^2 + 2x + 5 \) does not cross the \( x \)-axis at all.

This targets the confusion between “one solution” (discriminant = 0, where the graph just touches the axis) and “no real solutions” (discriminant < 0, where the graph misses the axis entirely). Students sometimes conflate these two cases because both feel like “something unusual” happening.

The student correctly identified \( a \), \( b \), \( c \) and correctly computed the discriminant as 1. The error is in the denominator: they divided by 2 instead of \( 2a = 2 \times 2 = 4 \). This is the “dividing by 2 instead of 2a” misconception. The formula has \( 2a \) in the denominator, not just 2.

The correct calculation is \( x = \frac{5 \pm 1}{4} \), giving \( x = \frac{6}{4} = 1.5 \) or \( x = \frac{4}{4} = 1 \). We can verify: \( 2(1.5)^2 \; – \; 5(1.5) + 3 = 4.5 \; – \; 7.5 + 3 = 0 \) ✓ and \( 2(1)^2 \; – \; 5(1) + 3 = 0 \) ✓. The student’s answers of 3 and 2 do not satisfy the equation: \( 2(3)^2 \; – \; 5(3) + 3 = 18 \; – \; 15 + 3 = 6 \neq 0 \).

The student has the correct answer (\( x = 1 \)), but their reasoning reveals the “incomplete formula” misconception — they have dropped the \( \pm\sqrt{b^2 \; – \; 4ac} \) term entirely and are treating \( x = \frac{-b}{2a} \) as the complete quadratic formula. This happened to give the right answer only because the discriminant here is \( 4 \; – \; 4 = 0 \), so the missing \( \pm\sqrt{0} \) contributes nothing.

For any equation where the discriminant is not zero, this shortcut fails. For example, applying \( x = \frac{-b}{2a} \) to \( x^2 \; – \; 4x + 3 = 0 \) gives \( x = 2 \), but the actual solutions are \( x = 1 \) and \( x = 3 \). Getting the right answer once does not mean the method is correct — it means the student got lucky with this particular equation.

The discriminant calculation is correct (\( (-3)^2 \; – \; 4(1)(-10) = 9 + 40 = 49 \) ✓), but the student has made a “sign error with −b”. Since \( b = -3 \), the formula gives \( -b = -(-3) = +3 \), so the numerator should be \( 3 \pm 7 \), not \( -3 \pm 7 \).

The correct calculation is \( x = \frac{3 \pm 7}{2} \), giving \( x = 5 \) or \( x = -2 \). Self-check strategy: You can always check the Sum of Roots. The sum should be \( -b/a \). Here, \( -(-3)/1 = 3 \). The student’s roots sum to \( 2 + (-5) = -3 \). Since \( -3 \neq 3 \), they could have spotted the error immediately!

The student has made the “forgetting to square b” error in the discriminant calculation. They wrote \( b^2 = 4 \), but \( b = 4 \), so \( b^2 = 16 \), not 4. The student has used the value of \( b \) itself (4) rather than \( b^2 \) (16). This is a common confusion between \( b \) and \( b^2 \), especially when \( b \) happens to be a small number.

The correct discriminant is \( 16 \; – \; 4 = 12 \). Since \( 12 > 0 \), there are two distinct solutions: \( x = \frac{-4 \pm \sqrt{12}}{2} = \frac{-4 \pm 2\sqrt{3}}{2} = -2 \pm \sqrt{3} \), approximately −0.27 and −3.73.

This is the “Heart-Shape Simplification” error. The student only divided one part of the numerator by the denominator. When simplifying a fraction like this, the denominator must divide both terms in the numerator.

The correct calculation is: \( \frac{8}{2} \pm \frac{4\sqrt{5}}{2} \). Since \( \frac{8}{2} = 4 \) AND \( \frac{4\sqrt{5}}{2} = 2\sqrt{5} \), the correct simplified answer is \( 4 \pm 2\sqrt{5} \).