KindiakMath

Quadratic Identities

Problem 1. Verify the following quadratic identities:

\begin{aligned} (a+b)^2 &= a^2 + 2ab + b^2, \\ (a-b)^2 &= a^2 - 2ab + b^2, \\ a^2 - b^2 &= (a+b)(a-b). \end{aligned}

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Solution. By the rainbow method,

\begin{aligned} (a+b)^2 &= (a+b)(a+b) \\ &= a(a+b) + b(a+b) \\ &= a^2 + ab + ba + b^2 \\ &= a^2 + ab + ab + b^2 \\ &= a^2 + 2ab + b^2. \end{aligned}

Replacing b with -b,

\begin{aligned} (a-b)^2 &= (a+(-b))^2 \\ &= a^2 + 2a(-b) + (-b)^2 \\ &= a^2 - 2ab + b^2. \end{aligned}

Finally,

\begin{aligned}(a+b)(a-b) &= a(a-b) + b(a-b) \\ &= a^2 - ab + ba - b^2 \\ &= a^2 - ab+ab - b^2 \\ &= a^2 - b^2.\end{aligned}

Let b, c be integers.

Problem 2. Define m := -b/2 and p:= c. Show that the quadratic equation

x^2 + bx + c = 0

has solutions given by x = m \pm \sqrt{m^2 - p}.

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Solution. Write b = -2m. Using the quadratic formula, the solutions are given by

\begin{aligned} x &= \frac{-b \pm \sqrt{b^2 - 4 \cdot 1 \cdot c}}{2 \cdot 1} \\ &= \frac{2m \pm \sqrt{(2m)^2 - 4p}}{2} \\ &= \frac{2m \pm \sqrt{4m^2 - 4p}}{2} \\ &= \frac{2m \pm \sqrt 4 \cdot \sqrt{m^2 - p}}{2} \\ &= \frac{2m \pm 2 \sqrt{m^2 - p}}{2} \\ &= m \pm \sqrt{m^2 - p}.\end{aligned}

Problem 3. Given that the quadratic equation

x^2 + bx + c = 0.

has roots r, s, evaluate b,c in terms of r,s. Deduce that

x^2 + bx + c = (x-r)(x-s),

and show that \Delta = 4(s-r)^2.

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Solution. Using Problem 2, suppose

r = m - \sqrt{m^2 - p},\quad s = m + \sqrt{m^2 - p}.

Then r+s = 2m = -b and by Problem 1,

\begin{aligned} rs &= (m - \sqrt{m^2 - p}) \cdot (m + \sqrt{m^2 - p})\\ &= m^2 - (m^2-p) \\ &= p = c. \end{aligned}

Therefore, b = -(r+s) and c = rs. Hence,

\begin{aligned} x^2 + bx + c &= x^2 -(r+s)x + rs \\ &= x^2 -rx-xs + rs \\ &= (x-r)x-(x-r)s \\ &= (x-r)(x-s). \end{aligned}

It follows that \Delta = 4 (m^2 - p) = 4(s-r)^2.

Now suppose c is a positive integer.

Problem 4. Suppose a quadratic equation of the form

x^2 - x - c = 0

has integer solutions with larger solution r. Evaluate c in terms of r.

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Solution. By Problem 3,

-(r+s) = -1\quad \Rightarrow \quad s = 1-r = -(r-1)

and c = -rs = -r(1-r) = r(r-1).

Problem 5. Show that a quadratic equation of the form

x^2 - x + c = 0

has no real roots.

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Solution. Since c \geq 1, by calculating the discriminant of the quadratic equation,

\begin{aligned} \Delta &= (-1)^2 - 4 \cdot 1 \cdot c \\ &= 1 - 4c \leq 1 - 4 \cdot 1 \\ &= -3 < 0. \end{aligned}

Therefore, the quadratic has no real roots.

Problem 6. Determine the values of r > 0 such that

x^2 - x + r = 0

has real roots.

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Solution. We require \Delta \geq 0:

(-1)^2 - 4 \cdot 1 \cdot r \geq 0 \quad \iff \quad r \leq 1/4.

Therefore, we require 0 < r \leq 1/4.

Problem 7. Given non-negative numbers a, b, show that

\displaystyle \sqrt{ab} \leq \frac{a+b}{2}.

This inequality is a special case of the arithmetic mean-geometric mean (AM-GM) inequality.

(Click for Solution)

Solution. Using Problem 1,

\begin{aligned} 0 &\leq (\sqrt{a}-\sqrt{b})^2 \\ &= (\sqrt{a})^2 - 2\sqrt{a}\sqrt{b} + (\sqrt{b})^2 \\ &= a - 2\sqrt{ab} + b. \end{aligned}

By algebruh, 2\sqrt{ab}\leq a+b. Dividing by 2 yields the desired result.

—Joel Kindiak, 24 Jan 26, 0213H

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