Quadratic Equations

Quadratic Equations

A quadratic equation can be solved by factoring, by completing the square, or with the quadratic formula. Factoring is fastest when the equation factors with integer coefficients. The quadratic formula, $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, solves any quadratic equation regardless of whether it factors. On the ACT, choose factoring first, and fall back on the formula whenever you cannot see the factors within a few seconds.

Solving by factoring

A quadratic factors cleanly when integer coefficients exist that multiply to give $ac$ and add to give $b$. Always set the equation equal to zero first, because the zero-product property only works when one side is $0$.

Worked Example. Solve $x^2 - 2x - 15 = 0$.

Find two numbers that multiply to $-15$ and add to $-2$. The numbers $-5$ and $3$ satisfy both conditions, so $(x - 5)(x + 3) = 0$ and $x = 5$ or $x = -3$.

When the leading coefficient is not $1$, as in $2x^2 - x - 15 = 0$, look for two numbers that multiply to $ac = -30$ and add to $-1$. Those numbers are $-6$ and $5$, which gives $(2x + 5)(x - 3) = 0$ and $x = -2.5$ or $x = 3$. A frequent wrong choice flips the sign of every root, so substitute one root back in to be sure.

The quadratic formula

When no integers work, use the formula. Identify $a$, $b$, and $c$ carefully, including their signs, then compute the discriminant $b^2 - 4ac$ first, since it feeds the rest of the formula. For $2x^2 - 3x - 5 = 0$, the discriminant is $9 + 40 = 49$, so $x = \frac{3 \pm 7}{4}$, which gives $2.5$ or $-1$. Remember that $-b$ means the opposite of $b$: if $b = -3$, then $-b = 3$.

The discriminant

The discriminant, $b^2 - 4ac$, determines the type of solutions a quadratic equation has before the equation is fully solved.

• A positive discriminant produces two real solutions.
• A discriminant of zero produces exactly one real solution.
• A negative discriminant produces two complex solutions.

Worked Example. Find the discriminant of $x^2 + 2x + 5 = 0$ and state the number of real solutions.

$(2)^2 - 4(1)(5) = 4 - 20 = -16$. The discriminant is negative, so the equation has zero real solutions and two complex solutions.

A negative discriminant does not stop the solving process; it signals that the two solutions will be complex conjugates, found the same way, with an imaginary term appearing under the radical. For $x^2 - 4x + 13 = 0$, $x = \frac{4 \pm \sqrt{16 - 52}}{2} = \frac{4 \pm 6i}{2} = 2 \pm 3i$.

Parameter questions

A parameter question asks for the value of an unknown coefficient that forces a specific discriminant condition, most often exactly one real solution. Set the discriminant equal to zero and solve for the parameter. For $x^2 + kx + 16 = 0$, the condition is $k^2 - 64 = 0$, so $k^2 = 64$ and $k = 8$ or $k = -8$. Do not forget the negative root; it is the standard trap.

The vertex and the sum and product of the roots

The vertex of a parabola $y = ax^2 + bx + c$ has $x$-coordinate $-\frac{b}{2a}$. Substitute that $x$-value into the equation to get the $y$-coordinate. For $y = x^2 - 8x + 15$, the $x$-coordinate is $4$ and the $y$-coordinate is $16 - 32 + 15 = -1$, so the vertex is $(4, -1)$. The vertex is also the minimum point when $a > 0$ and the maximum point when $a < 0$.

For any quadratic $ax^2 + bx + c = 0$, the sum of the solutions equals $-\frac{b}{a}$ and the product of the solutions equals $\frac{c}{a}$, without solving the equation directly. For $x^2 - 7x + 10 = 0$ the sum is $7$ and the product is $10$ (the roots are $2$ and $5$). These formulas hold even when the roots are complex.

Where this appears as a model

An object's height over time is a quadratic model. An object launched from $5$ feet with an initial upward speed of $64$ feet per second follows $h(t) = -16t^2 + 64t + 5$. The maximum height is found at the vertex: $t = -\frac{64}{2(-16)} = 2$ seconds, and the height is $-16(4) + 128 + 5 = 69$ feet.

Strategy checklist

• Write the equation as $ax^2 + bx + c = 0$ before doing anything else.
• Try to factor; if nothing works quickly, use the formula.
• For "how many real solutions" questions, compute only the discriminant.
• For "exactly one solution" parameter questions, set the discriminant to $0$ and keep both roots.
• For vertex and sum/product questions, use $-\frac{b}{2a}$, $-\frac{b}{a}$, and $\frac{c}{a}$ instead of solving.