The Quadratic Formula: Practice

In standard form $ax^2 + bx + c = 0$, the number $a$ multiplies $x^2$, $b$ multiplies $x$, and $c$ is the constant, each read with its sign.

$a = 1,\ b = -7,\ c = 10$

The middle term $-7x$ carries a minus sign, so $b = -7$, and the coefficient of $x^2$ is $1$.

The formula divides the entire numerator by $2a$.

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

The numerator begins with $-b$, the denominator is $2a$ (not $2$), and the whole top is over $2a$, not just the square-root term.

With $a = 1$, $b = 6$, $c = 5$, the discriminant is $b^2 - 4ac$.

$D = 6^2 - 4(1)(5) = 36 - 20 = 16$

Subtract $4ac$ from $b^2$; adding them would wrongly give $56$.

A zero discriminant means $\sqrt{D} = 0$, so the $\pm$ adds nothing and the two roots coincide.

$D = (-4)^2 - 4(1)(4) = 16 - 16 = 0$

There is exactly one repeated real root, $x = -\frac{b}{2a} = 2$.

A square root of a negative number uses the imaginary unit: $\sqrt{-n} = i\sqrt{n}$.

$\sqrt{-25} = i\sqrt{25} = 5i$

No real number squares to $-25$, so the answer must carry an $i$.

The value $b$ is the coefficient of the $x$ term, read with its sign. The term is $-x = -1x$.

$b = -1$

The $5$ is $a$ (the coefficient of $x^2$) and the $4$ is $c$ (the constant).

A negative discriminant means $\sqrt{D}$ is imaginary.

$\sqrt{-3} = i\sqrt{3}$

The $\pm$ then produces a conjugate pair, so there are two complex roots. Complex roots of a real quadratic always come in conjugate pairs, so a mix of one real and one complex root cannot happen.

Read off $a = 1$, $b = 7$, $c = 12$, so $D = 49 - 48 = 1$.

$x = \frac{-7 \pm \sqrt{1}}{2} = \frac{-7 \pm 1}{2}$

So $x = \frac{-6}{2} = -3$ or $x = \frac{-8}{2} = -4$.

The discriminant is the quantity sitting under the square root.

$D = b^2 - 4ac$

Its sign (positive, zero, or negative) decides whether the roots are two real, one repeated, or complex.

The formula needs standard form, with everything on one side equal to $0$. Subtract $4x$ from both sides.

$x^2 + 5 = 4x \quad\Rightarrow\quad x^2 - 4x + 5 = 0$

Only then are $a = 1$, $b = -4$, $c = 5$ correct.

Here $b = -8$, and the formula uses $-b$, the opposite of $b$.

$-b = -(-8) = 8$

So the numerator begins with $+8$, not $-8$.

Compute the discriminant with $a = 3$, $b = -2$, $c = 5$.

$D = (-2)^2 - 4(3)(5) = 4 - 60 = -56$

A negative discriminant means both roots are complex, so there are no real roots. (The equation still has two roots, just not real ones.)