Quadratic Functions and Equations

Express $2x^2 - 8x + 5$ in the form $a(x + h)^2 + k$.

$2x^2 - 8x + 5 = 2(x^2 - 4x) + 5$

$= 2(x^2 - 4x + 4 - 4) + 5$

$= 2(x - 2)^2 - 8 + 5 = 2(x - 2)^2 - 3$

Vertex: $(2, -3)$. Since $a = 2 > 0$, this is a minimum.

Questions asking for the "range" of a quadratic almost always require completing the square first to identify $k$. Quote the range as an inequality: $f(x) \geq k$ or $f(x) \leq k$.

Determine the values of $k$ for which $kx^2 + (k-3)x + 1 = 0$ has no real roots.

For no real roots: $\Delta < 0$.

$\Delta = (k-3)^2 - 4k(1) = k^2 - 6k + 9 - 4k = k^2 - 10k + 9$

Set $\Delta < 0$: $(k-1)(k-9) < 0$, so $1 < k < 9$.

Also, for the equation to be quadratic, $k \neq 0$. Since $k = 0$ is outside the interval $1 < k < 9$, the final answer is $1 < k < 9$.

When a question says "solve," any valid method is acceptable. When it says "complete the square" or "express in the form $a(x+h)^2 + k$," you must use that specific method.

Solve $x - 5\sqrt{x} + 6 = 0$.

Let $u = \sqrt{x}$ (so $u \geq 0$ and $x = u^2$):

$u^2 - 5u + 6 = 0 \Rightarrow (u-2)(u-3) = 0$

so $u = 2$ or $u = 3$. Since $u = \sqrt{x}$: $x = 4$ or $x = 9$.

Check: $4 - 5(2) + 6 = 0$ ✓ and $9 - 5(3) + 6 = 0$ ✓

An extraneous solution is a value produced by the algebra that does not satisfy the original equation. It arises because a substitution (like $u = \sqrt{x}$, which requires $u \geq 0$) imposes a domain restriction that the factored quadratic silently ignores. Always verify every back-substituted solution in the original equation. Any value that fails the check must be marked (Invalid) and discarded — the algebra will not warn you automatically.

The roots of $3x^2 - 5x + 2 = 0$ are $\alpha$ and $\beta$. Find $\alpha^2 + \beta^2$.

$\alpha + \beta = \frac{5}{3}, \quad \alpha\beta = \frac{2}{3}$

$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = \frac{25}{9} - \frac{4}{3} = \frac{25}{9} - \frac{12}{9} = \frac{13}{9}$

The roots of $2x^2 - 3x + 1 = 0$ are $\alpha$ and $\beta$. Find the quadratic with integer coefficients whose roots are $\alpha^2$ and $\beta^2$.

From the original:

$\alpha + \beta = \frac{3}{2}$

$\alpha\beta = \frac{1}{2}$

New sum:

$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = \frac{9}{4} - 1 = \frac{5}{4}$

New product:

$\alpha^2\beta^2 = (\alpha\beta)^2 = \frac{1}{4}$

Substituting into the formula:

$x^2 - \frac{5}{4}x + \frac{1}{4} = 0$

Coefficients are not integers. Multiply through by 4:

$4x^2 - 5x + 1 = 0$

Solve the system: $y = x + 1$ and $x^2 + y^2 = 13$.

Substitute $y = x + 1$ into the circle:

$x^2 + (x+1)^2 = 13$

$x^2 + x^2 + 2x + 1 = 13$

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

$x = -3$ gives $y = -2$; $x = 2$ gives $y = 3$.

Solutions: $(-3, -2)$ and $(2, 3)$.

When a line intersects a circle, the discriminant of the resulting quadratic tells you about the intersection. $\Delta > 0$: two points; $\Delta = 0$: the line is a tangent; $\Delta < 0$: no intersection.