Quadratic Equations: questions with answers

A quadratic polynomial has at most two zeroes, and the roots of $ax^2 + bx + c = 0$ are exactly its zeroes. So a quadratic equation has at most two roots. This leaves room for two equal roots, and for $x^2 + x + 1 = 0$, which has no real root at all.

A quadratic polynomial has at most two zeroes, which sets an upper limit and not a fixed count. $x^2 + x + 1 = 0$ has discriminant $1 - 4 = -3$, so it has no real root at all. $x^2 - 2x + 1 = 0$ has discriminant $0$, so it has only one distinct real root. Both cases show the equation does not always have exactly two.

If the roots are $\alpha$ and $-\alpha$ their sum is zero. The sum is $-\frac{b}{a}$, so $-\frac{b}{a} = 0$ and therefore $b = 0$. Nothing is forced on $c$: the product is $-\alpha^2$, which is negative but otherwise free. And $a$ cannot be zero at all, or the equation would not be quadratic.

Each factor is set to zero and then SOLVED. $2x - 1 = 0$ gives $x = \frac{ 1 }{ 2 }$, not $1$ -- a factor carrying a coefficient on $x$ still has a division left in it. The second factor, $x - 2 = 0$, gives $x = 2$ directly.

Multiplying an equation means multiplying EVERY term on both sides. $\frac{ 6 }{x} \times x = 6$, $-3 \times x = -3x$ and $7 \times x = 7x$, which collects to $10x - 6 = 0$. Leaving the whole-number term untouched produces a different equation, and so a different answer.

Splitting the middle term gives $(x + 3)(x - 7) = 0$, so the roots are $-3$ and $7$. Each factor is SET TO ZERO and solved, which reverses the sign printed inside it: the bracket $(x + 3)$ yields the root $-3$, not $3$.

In standard form $a = -3$, $b = 7$ and $c = 8$, so $a - b + c = -3 - 7 + 8$. Reading the coefficients in printed order instead gives $a = 7$ and $b = -3$. This wrong reading gives $7 + 3 + 8$, a different number, since subtracting $b$ is not symmetric.

The expanded sides are $x^2 + 6x + 9$ and $x^2 + 3x - 12$. Subtracting the right side cancels the squared terms and gives $3x + 21 = 0$. Omitting the middle term of the square gives $-3x + 21 = 0$. Retaining a squared term instead of cancelling gives $x^2 + 3x + 21 = 0$.

The left side expands to $x^3 + 9x^2 + 27x + 27$. The $x^3$ on each side cancels, leaving $9x^2 + 27x + 23 = 0$, whose highest power is $2$. So the degree of the simplified equation is $2$, not $3$ -- the equation only looks cubic until the $x^3$ terms are collected.

Time is distance over speed, so the actual journey takes $\frac{ 480 }{x}$ hours and the faster one $\frac{ 480 }{x + 10}$ hours. A greater speed takes LESS time, so the faster time is the smaller of the two and the hour saved is the slower time minus the faster: $\frac{ 480 }{x} - \frac{ 480 }{x + 10}$. Subtracting the other way round gives a negative number of hours saved, and adding gives the total time for two journeys.

Read the sentence in the order it is built: start with the breadth, multiply it by 3, then add 1. That is $3x + 1$. Writing $1x + 3$ swaps the two numbers, giving a different plot. And $3(x + 1)$ adds first and multiplies afterwards, so the 1 m also gets multiplied by 3.

Expanding first gives $kx^2 - 3kx + 8 = 0$, so $a = k$, $b = -3k$ and $c = 8$. The letter multiplies the whole bracket, so it reaches the linear term as well as the square. Reading $b = -3$ off the printed form leaves the $k$ behind, and every discriminant computed from it is then wrong.

The condition $b^2 - 4ac = 0$ reads $1 - 4a^2 = 0$, so $a = \pm\frac{ 1 }{2}$ -- both values give equal roots. The extra word decides between them: the repeated root is $-\frac{ 1 }{2a}$, which is positive only when $a$ is negative. So $a = -\frac{ 1 }{2}$, and the positive value is thrown out even though it satisfies the discriminant.

Equal roots mean the discriminant is zero: $b^2 - 4ac = 0$, that is $b^2 = 4ac$. The $4$ is part of the discriminant and cannot be dropped -- $b^2 = ac$ would be the condition for a different equation entirely. Nor may the subtraction become an addition: $b^2 = -4ac$ describes when the discriminant $b^2 + 4ac$ vanishes, which is not the quantity in the formula.

Here $a = 1$, $b = -4$ and $c = -5$. The constant is NEGATIVE, so $-4ac = -4(1)(-5) = +20$ and the discriminant is $4^2 + 20 = 36$. Dropping the minus on $c$ turns that addition back into a subtraction and gives $-4$ instead. Two minus signs meet in that term, so losing one makes the discriminant shrink instead of grow.