Quadratic Equations: Chapter Test

A product is zero only when one of its factors is zero, so each factor gives its own linear equation.

$x + 8 = 0 \quad \text{or} \quad x - 3 = 0$

Solving each gives $x = -8$ or $x = 3$. Each root is the value that makes its own factor vanish, which is the opposite of the number written inside it.

Substitute each candidate and look for a left side of exactly $0$.

$4^2 - 4 - 12 = 16 - 4 - 12 = 0$

So $4$ satisfies the equation. Testing $-4$ gives $16 + 4 - 12 = 8$, testing $3$ gives $9 - 3 - 12 = -6$, and testing $12$ gives $144 - 12 - 12 = 120$, none of them $0$. (The second solution is $-3$, which is not offered.)

Whatever is inside the square is one quantity, and that quantity has to be one of the two numbers whose square is $81$.

$x + 7 = \pm 9 \quad\Longrightarrow\quad x = -7 + 9 = 2 \quad \text{or} \quad x = -7 - 9 = -16$

Both check: $(2 + 7)^2 = 81$ and $(-16 + 7)^2 = (-9)^2 = 81$.

The constant $28$ is positive, so the two numbers share a sign, and the middle coefficient $-11$ is negative, so both are negative. Among the negative pairs multiplying to $28$, only $-4$ and $-7$ add to $-11$.

$x^2 - 11x + 28 = (x - 4)(x - 7)$

Expanding returns the original: $(x - 4)(x - 7) = x^2 - 7x - 4x + 28 = x^2 - 11x + 28$.

For a monic quadratic the roots add to $-b$ and multiply to $c$. Here $b = -3$ and $c = -28$, so the double negative on the sum turns into a plus.

$r + s = -b = -(-3) = 3, \qquad rs = c = -28$

As a check, this trinomial factors as $(x - 7)(x + 4)$, and its roots $7$ and $-4$ do add to $3$ and multiply to $-28$.

Isolate the square in each and look at what is left on the other side. A real number squared is never negative, so only a negative right-hand side is impossible.

$(x + 3)^2 = -4 \quad \text{is impossible}, \qquad x^2 = 4, \qquad (x - 5)^2 = 0, \qquad x^2 = 16 \quad \text{are not}$

The other three have two, one, and two real solutions in turn.

The middle term splits into two numbers with sum $-9$ and product $a \cdot c = 10 \times 2 = 20$, which are $-4$ and $-5$. Pairing $2x$ with $-2$ makes a $-4x$, and pairing $5x$ with $-1$ makes a $-5x$.

$(2x - 1)(5x - 2) = 10x^2 - 4x - 5x + 2 = 10x^2 - 9x + 2$

The other three expand to $10x^2 - 12x + 2$, $10x^2 - 21x + 2$, and $10x^2 + 9x + 2$.

A product splits into its factors only when the product is $0$, because $0$ is the one number that no two nonzero numbers can multiply to. With $56$ on the right, nothing follows. Expand and gather everything onto one side first.

$x^2 + 3x + 2 = 56 \quad\Longrightarrow\quad x^2 + 3x - 54 = 0 \quad\Longrightarrow\quad (x + 9)(x - 6) = 0$

So $x = -9$ or $x = 6$, and both check: $(-8)(-7) = 56$ and $(7)(8) = 56$.

Every term has to be on one side, with $0$ on the other, before a single coefficient can be read. Subtracting $2x$ and $8$ from both sides collects $-x - 2x = -3x$.

$3x^2 - x = 2x + 8 \quad\Longrightarrow\quad 3x^2 - 3x - 8 = 0$

So $a = 3$, $b = -3$, and $c = -8$.

The two totals are the only things the coefficients report. Two negative numbers add to a negative number and multiply to a positive one.

$r + s = -b < 0 \quad\Longrightarrow\quad b > 0, \qquad rs = c > 0$

So $b$ and $c$ are both positive. The trinomial $x^2 + 7x + 12 = (x + 3)(x + 4)$ is one instance, with roots $-3$ and $-4$.

All three coefficients are multiples of $3$, so pull it out first; inside, the leading coefficient is $1$ and two numbers with sum $2$ and product $-35$ are $7$ and $-5$.

$3x^2 + 6x - 105 = 3\left(x^2 + 2x - 35\right) = 3(x + 7)(x - 5)$

The first option does rebuild the original on expansion, but $3x + 21$ is itself $3(x + 7)$, so a factor is still hidden inside and the factoring is not finished.

Put the equation in standard form as $x^2 + 0x - 40 = 0$ and read the two totals off the coefficients, with no need to evaluate a root at all.

$r + s = -b = 0, \qquad rs = c = -40$

That matches the solutions themselves, $\sqrt{40}$ and $-\sqrt{40}$: opposites add to $0$, and a positive times a negative is negative.

The middle term splits into two numbers with sum $1$ and product $a \cdot c = 6 \times (-12) = -72$, which are $9$ and $-8$. That builds the factors, and expanding confirms them.

$6x^2 + x - 12 = (3x - 4)(2x + 3) = 0$

Setting each factor to zero gives $3x = 4$, so $x = \tfrac{4}{3}$, and $2x = -3$, so $x = -\tfrac{3}{2}$. The fractions come from dividing out the numbers in front of $x$.

Nothing can be split while the right side is not $0$, so expand first and gather every term onto one side. Both the $2x$ and the $-15$ cancel.

$x^2 + 2x - 15 = 2x - 15 \quad\Longrightarrow\quad x^2 = 0$

The only number whose square is $0$ is $0$ itself, so there is a single solution, $x = 0$. Checking it against the original: $(-3)(5) = -15$ and $2(0) - 15 = -15$.

Call the repeated root $r$, so the two roots are $r$ and $r$. Their product is the constant term, which pins $r$ down, and a square has two roots.

$r \cdot r = 36 \quad\Longrightarrow\quad r = 6 \text{ or } r = -6$

Their sum is $-b$, so $2r = -b$ gives $b = -12$ when $r = 6$ and $b = 12$ when $r = -6$. Both work: $x^2 - 12x + 36 = (x - 6)^2$ and $x^2 + 12x + 36 = (x + 6)^2$.

Gather every term onto one side first, so that a product can be set against $0$. Subtracting $6x$ and $25$ leaves a monic trinomial.

$x^2 - 4x - 45 = 0 \quad\Longrightarrow\quad (x - 9)(x + 5) = 0$

Two numbers with sum $-4$ and product $-45$ are $-9$ and $5$, and each factor then gives its root with the sign flipped: $x = 9$ or $x = -5$.

Expand each candidate and read only its middle term, since all four already give $12x^2$ and $+12$. The middle term is the sum of the two cross products.

$(4x - 3)(3x - 4) = 12x^2 - 16x - 9x + 12 = 12x^2 - 25x + 12$

The others give $-24x$, $-30x$, and $-51x$. The aimed search agrees: two numbers with sum $-25$ and product $a \cdot c = 144$ are $-16$ and $-9$, exactly the two cross products above.

The search asked one narrow question: is there a pair of INTEGERS with sum $5$ and product $5$? The pairs are $1$ and $5$, adding to $6$, and $-1$ and $-5$, adding to $-6$, so the answer is no.

$x^2 + 5x + 5 \text{ has no integer factor pair} \quad\Longrightarrow\quad \text{it is irreducible over the integers}$

That is a statement about factoring over the integers, and nothing more. It rules out no non-integer solution, because the search never examined one, and forcing $-1$ and $-5$ through anyway fails at once, since $(-1)^2 + 5(-1) + 5 = 1$. Tools that settle such an equation come later.

The right side is not $0$, so expand the left and gather everything onto one side before splitting anything.

$6x^2 + 10x - 4 = 5x + 2 \quad\Longrightarrow\quad 6x^2 + 5x - 6 = 0$

Now the middle term splits into two numbers with sum $5$ and product $6 \times (-6) = -36$, namely $9$ and $-4$, which build $(3x - 2)(2x + 3)$. Expanding confirms it, and setting each factor to zero gives $x = \tfrac{2}{3}$ or $x = -\tfrac{3}{2}$.

Combine the two reciprocals over a common denominator first, and what appears is exactly the sum and the product of the roots.

$\frac{1}{r} + \frac{1}{s} = \frac{s + r}{rs}$

The coefficients hand both totals over without solving: $r + s = -\tfrac{-10}{2} = 5$ and $rs = \tfrac{3}{2}$. Dividing gives $5 \div \tfrac{3}{2} = 5 \cdot \tfrac{2}{3} = \tfrac{10}{3}$.