Quadratic Equations — Balbharati Solutions, AI Tutor & Practice

A quadratic equation has the standard form $ax^2 + bx + c = 0$ where $a \neq 0$.

(i) $x^2 + 5x - 3 = 0$: Here the highest power of $x$ is $2$ and $a = 1 \neq 0$. It is a quadratic equation.

(ii) $x + \frac{1}{x} = 4$. Multiply both sides by $x$: $x^2 + 1 = 4x$, i.e. $x^2 - 4x + 1 = 0$. Highest power is $2$. It is a quadratic equation.

(iii) $2x - 7 = 0$: highest power of $x$ is $1$. It is NOT a quadratic equation (it is linear).

A value is a root if it satisfies the equation, i.e. makes LHS $=$ RHS.

Substitute $x = -2$ into $x^2 + 3x + 2$: $(-2)^2 + 3(-2) + 2 = 4 - 6 + 2 = 0.$

Since LHS $= 0 =$ RHS, $x = -2$ is a root of the equation.

We need two numbers with product $12$ and sum $-7$. These are $-3$ and $-4$.

Split the middle term: $x^2 - 3x - 4x + 12 = 0$ $x(x - 3) - 4(x - 3) = 0$ $(x - 3)(x - 4) = 0$

So $x - 3 = 0$ or $x - 4 = 0$, giving $x = 3 \quad \text{or} \quad x = 4.$

Roots: $3$ and $4$.

Start with $x^2 + 6x - 7 = 0$, so $x^2 + 6x = 7$.

Half the coefficient of $x$ is $3$; its square is $9$. Add $9$ to both sides: $x^2 + 6x + 9 = 7 + 9$ $(x + 3)^2 = 16$

Take square roots: $x + 3 = \pm 4$ $x = -3 + 4 = 1 \quad \text{or} \quad x = -3 - 4 = -7.$

Roots: $1$ and $-7$.

Here $a = 2$, $b = -5$, $c = -3$.

Discriminant: $b^2 - 4ac = (-5)^2 - 4(2)(-3) = 25 + 24 = 49.$

Apply the formula: $x = \frac{-(-5) \pm \sqrt{49}}{2(2)} = \frac{5 \pm 7}{4}.$

So $x = \frac{5 + 7}{4} = 3 \quad \text{or} \quad x = \frac{5 - 7}{4} = -\frac{1}{2}.$

Roots: $3$ and $-\dfrac{1}{2}$.

Here $a = 3$, $b = -2$, $c = 1$.

Discriminant: $\Delta = b^2 - 4ac = (-2)^2 - 4(3)(1) = 4 - 12 = -8.$

Since $\Delta = -8 < 0$, the equation has no real roots (the roots are not real / are imaginary).

For $x^2 - kx + 12 = 0$: sum of roots $= k$, product of roots $= 12$.

Let the roots be $3$ and $\beta$.

Product: $3 \times \beta = 12 \Rightarrow \beta = 4$.

Sum: $3 + \beta = k \Rightarrow k = 3 + 4 = 7$.

Therefore $k = 7$ and the other root is $4$.

Check: substituting $x = 3$: $9 - 7(3) + 12 = 9 - 21 + 12 = 0$. ✓

Let $\alpha = 5$, $\beta = -2$.

Sum of roots: $\alpha + \beta = 5 + (-2) = 3$.

Product of roots: $\alpha\beta = 5 \times (-2) = -10$.

The required equation is $x^2 - (\alpha + \beta)x + \alpha\beta = 0$ $x^2 - 3x - 10 = 0.$

Roots are real and equal when the discriminant is zero: $b^2 - 4ac = 0$.

Here $a = 1$, $b = k$, $c = 9$: $k^2 - 4(1)(9) = 0$ $k^2 - 36 = 0$ $k^2 = 36$ $k = \pm 6.$

The roots are equal when $k = 6$ or $k = -6$.

Let the first even integer be $x$. The next consecutive even integer is $x + 2$.

According to the condition: $x(x + 2) = 168$ $x^2 + 2x - 168 = 0.$

Factorise (product $-168$, sum $+2$ gives $14$ and $-12$): $x^2 + 14x - 12x - 168 = 0$ $x(x + 14) - 12(x + 14) = 0$ $(x + 14)(x - 12) = 0.$

So $x = -14$ or $x = 12$. Since the integers are positive, $x = 12$.

The integers are $12$ and $14$.

Check: $12 \times 14 = 168$. ✓

Let the number be $x$. Its reciprocal is $\frac{1}{x}$.

Given: $x + \frac{1}{x} = \frac{13}{6}.$

Multiply throughout by $6x$: $6x^2 + 6 = 13x$ $6x^2 - 13x + 6 = 0.$

Factorise (product $36$, sum $-13$ gives $-9$ and $-4$): $6x^2 - 9x - 4x + 6 = 0$ $3x(2x - 3) - 2(2x - 3) = 0$ $(2x - 3)(3x - 2) = 0.$

So $x = \dfrac{3}{2}$ or $x = \dfrac{2}{3}$.

The number is $\dfrac{3}{2}$ or $\dfrac{2}{3}$. (These are reciprocals of each other, consistent with the problem.)

Let the speed of the train be $x$ km/h.

Time at speed $x$: $\dfrac{360}{x}$ hours. Time at speed $(x+5)$: $\dfrac{360}{x+5}$ hours.

The faster trip takes $1$ hour less: $\frac{360}{x} - \frac{360}{x+5} = 1.$

Combine the fractions: $360\left(\frac{(x+5) - x}{x(x+5)}\right) = 1$ $\frac{360 \times 5}{x(x+5)} = 1$ $1800 = x^2 + 5x$ $x^2 + 5x - 1800 = 0.$

Factorise (product $-1800$, sum $+5$ gives $45$ and $-40$): $(x + 45)(x - 40) = 0.$

So $x = -45$ or $x = 40$. Speed cannot be negative, so $x = 40$.

The speed of the train is $40$ km/h.