Algebra - Equations and Inequations

The roots $x_1$ and $x_2$ of the equation $x^2 + px + 12 = 0$ are such that $x_2 - x_1 = 1$. Find $p$.

Given quadratic equation: $\;\;\;$ $x^2 + px + 12 = 0$

Relation between the roots $x_1$ and $x_2$ of the given quadratic equation is

$x_2 - x_1 = 1$ $\;\;\; \cdots \; (1)$

Sum of roots of the given quadratic equation: $\;$ $x_1 + x_2 = -p$ $\;\;\; \cdots \; (2)$

Adding equations $(1)$ and $(2)$ simultaneously gives

$2 x_2 = 1 - p$ $\implies$ $x_2 = \dfrac{1-p}{2}$

Substituting the value of $x_2$ in equation $(1)$ gives

$x_1 = x_2 - 1 = \dfrac{1 - p}{2} - 1 = - \left(\dfrac{1 + p}{2}\right)$

Product of roots of the given quadratic equation: $\;$ $x_1 \cdot x_2 = 12$ $\;\;\; \cdots \; (3)$

Substituting the values of $x_1$ and $x_2$ in equation $(3)$ gives

$- \left(\dfrac{1 - p}{2}\right) \left(\dfrac{1 + p}{2}\right) = 12$

i.e. $\;$ $1 - p^2 = - 48$

i.e. $p^2 = 49$

i.e. $\;$ $p = \pm 7$

Algebra - Equations and Inequations

Find $\;$ $a$ $\;$ such that one of the roots of the equation $\;$ $x^2 - \dfrac{15}{4} x + a = 0$ $\;$ is the square of the other.

Given quadratic equation: $\;\;\;$ $x^2 - \dfrac{15}{4} x + a = 0$ $\;\;\; \cdots \; (1)$

Given: $\;$ One root of equation $(1)$ is the square of the other.

Let the roots of $(1)$ be $\alpha, \; \alpha^2$.

Sum of roots of equation $(1)$ $= \alpha + \alpha^2 = \dfrac{- \left(-15/4\right)}{1} = \dfrac{15}{4}$

i.e. $\;$ $\alpha^2 + \alpha - \dfrac{15}{4} = 0$

i.e. $\;$ $4 \alpha^2 + 4 \alpha - 15 = 0$

i.e. $\;$ $\left(2 \alpha + 5\right) \left(2 \alpha - 3\right) = 0$

i.e. $\;$ $\alpha = \dfrac{-5}{2}$, $\;\;$ or $\;\;$ $\alpha = \dfrac{3}{2}$

Product of roots of equation $(1)$ $= \alpha \times \alpha^2 = \alpha^3 = \dfrac{a}{1} = a$ $\;\;\; \cdots \; (2)$

Substituting $\alpha = \dfrac{-5}{2}$ in equation $(2)$ gives

$a = \left(\dfrac{-5}{2}\right)^3 = \dfrac{-125}{8}$

Substituting $\alpha = \dfrac{3}{2}$ in equation $(2)$ gives

$a = \left(\dfrac{3}{2}\right)^3 = \dfrac{27}{8}$

Algebra - Equations and Inequations

Find the value of $\;$ $a$ $\;$ for which one root of the equation $\;$ $x^2 + \left(2a-1\right)x + a^2 + 2 = 0$ $\;$ is twice as large as the other.

Given quadratic equation: $\;\;\;$ $x^2 + \left(2a-1\right)x + a^2 + 2 = 0$ $\;\;\; \cdots \; (1)$

Given: $\;$ One root of equation $(1)$ is twice as large as the other.

Let the roots of $(1)$ be $\alpha, \; 2 \alpha$.

Sum of roots of equation $(1)$ $= \alpha + 2 \alpha = 3 \alpha = \dfrac{- \left(2a - 1\right)}{1} = 1 - 2a$

$\implies$ $\alpha = \dfrac{1 - 2a}{3}$ $\;\;\; \cdots \; (2)$

Product of roots of equation $(1)$ $= \alpha \times 2 \alpha = 2 \alpha^2 = \dfrac{a^2 + 2}{1}$

i.e. $\;$ $2 \alpha^2 = a^2 + 2$ $\;\;\; \cdots \; (3)$

In view of equation $(2)$, equation $(3)$ becomes

$2 \times \left(\dfrac{1 - 2a}{3}\right)^2 = a^2 + 2$

i.e. $\;$ $\dfrac{2 \left(1 - 4a + 4a^2\right)}{9} = a^2 + 2$

i.e. $\;$ $2 - 8a + 8a^2 = 9a^2 + 18$

i.e. $\;$ $a^2 + 8a + 16 = 0$

i.e. $\;$ $\left(a + 4\right)^2 = 0$

$\implies$ $a = -4$

Algebra - Equations and Inequations

For what values of $a$ is the ratio of the roots of the equation $ax^2 - \left(a + 3\right)x + 3 = 0$ equal to $1.5$?

Given quadratic equation: $\;\;\;$ $ax^2 - \left(a + 3\right)x + 3 = 0$ $\;\;\; \cdots \; (1)$

Comparing equation $(1)$ with the standard quadratic equation $\;$ $Ax^2 + Bx + C = 0$ $\;$ gives

$A = a, \; B = - \left(a + 3\right), \; C = 3$

Roots of equation $(1)$ are $x = \dfrac{a + 3 \pm \sqrt{\left[-\left(a + 3\right)\right]^2 - 4 \times a \times 3}}{2 \times a}$

i.e. $\;$ $x = \dfrac{a + 3 \pm \sqrt{a^2 + 6a + 9 - 12a}}{2a}$

i.e. $\;$ $x = \dfrac{a + 3 \pm \sqrt{a^2 - 6a + 9}}{2a}$

i.e. $\;$ $x = \dfrac{a + 3 \pm \sqrt{\left(a - 3\right)^2}}{2a}$

i.e. $\;$ $x = \dfrac{a + 3 \pm \left(a - 3\right)}{2a}$

$\therefore \;$ The roots are $\;\;$ $x_1 = \dfrac{a + 3 + a - 3}{2a} = 1$, $\;$ $x_2 = \dfrac{a + 3 - a + 3}{2a} = \dfrac{3}{a}$

As per question, ratio of roots $= 1.5$

i.e. $\;$ $\dfrac{x_1}{x_2} = \dfrac{1}{3/a} = 1.5$ $\implies$ $a = 4.5$

or $\;$ $\dfrac{x_2}{x_1} = 1.5 = \dfrac{3/a}{1} = 1.5$ $\implies$ $a = 2$

Algebra - Equations and Inequations

For what values of $a$ is the sum of the roots of the equation $x^2 + \left(2-a-a^2\right)x - a^2 = 0$ equal to zero?

Given quadratic equation: $\;\;\;$ $x^2 + \left(2-a-a^2\right)x - a^2 = 0$ $\;\;\; \cdots \; (1)$

Let the roots of $(1)$ be $\alpha$ and $\beta$

Sum of roots $= \alpha + \beta = \dfrac{-\left(2 - a - a^2\right)}{1} = a^2 + a - 2$

as per question, $\;$ $\alpha + \beta = 0$

i.e. $\;$ $a^2 + a - 2 = 0$

i.e. $\;$ $\left(a + 2\right) \left(a - 1\right)= 0$

i.e. $\;$ $a = -2$ $\;$ or $\;$ $a = 1$

Algebra - Equations and Inequations

Find the least integral value of $k$ for which the equation $x^2 - 2 \left(k + 2\right)x + 12 + k^2 = 0$ has two different real roots.

Given quadratic equation: $\;\;\;$ $x^2 - 2 \left(k + 2\right) x + 12 + k^2 = 0$ $\;\;\; \cdots \; (1)$

Equation $(1)$ has two different real roots when its discriminant $\Delta > 0$

i.e. $\;$ $\Delta = \left[-2 \left(k + 2\right)\right]^2 - 4 \times 1 \times \left(12 + k^2\right) > 0$

i.e. $\;$ $4 \left(k^2 + 4k + 4\right) - 48 - 4k^2 >0$

i.e. $\;$ $4k^2 + 16k + 16 - 48 - 4k^2 > 0$

i.e. $\;$ $16 k > 32$

i.e. $\;$ $k > 2$

$\therefore \;$ The least value of $k$ for which equation $(1)$ has two different real roots is $k = 3$

Algebra - Equations and Inequations

For what values of $k$ is the inequality $x^2 - \left(k-3\right) x - k + 6 > 0$ valid for all real $x$?

Consider the quadratic equation $\;\;$ $x^2 - \left(k-3\right)x - k + 6 = 0$ $\;\;\; \cdots \; (1)$

Equation $(1)$ will have real roots when its discriminant

$\Delta = \left[- \left(k - 3\right)\right]^2 - 4 \times 1 \times \left(6 - k\right) > 0$

i.e. $\;$ $k^2 - 6k + 9 - 24 + 4k > 0$

i.e. $\;$ $k^2 - 2k - 15 > 0$

i.e. $\;$ $\left(k + 3\right) \left(k - 5\right) > 0$

i.e. $\;$ $k + 3 > 0$ $\;$ and $\;$ $k - 5 > 0$ $\;\;$ OR $\;\;$ $k + 3 < 0$ $\;$ and $\;$ $k - 5 < 0$

i.e. $\;$ $k > -3$ $\;$ and $\;$ $k > 5$ $\;\;$ OR $\;\;$ $k < -3$ $\;$ and $\;$ $k < 5$

i.e. $\;$ $k > 5$ $\;\;$ OR $\;\;$ $k < -3$

i.e. $\;$ $5 < k < -3$

$\therefore \;$ The given inequality is valid for all real $x$ when $k \in \left(-3, 5\right)$