**1.** Discuss the nature of the roots of $x^2 + 5x + 4 = 0.$

Solution 1:

Given: $x^2 + 5x + 4 = 0$

Here, $a = 1, b = 5, c = 4$

$\therefore \Delta = b^2 – 4ac = 9$

Thus, D = 9 is a perfect square.

Hence, the roots are real, rational and unequal.

**2.** Prove that for $ax^2 + by^2 = 1$ and $ax + by = 1,$ the values of $x$ will be equal if $a + b = 1.$

Solution 2:

Given, $ax + by = 1$

i.e. $y = \frac{1 – ax}{b}$

Again, $ax^2 + by^2 = 1$

$\therefore$ $ax^2 + \frac{(1 – ax)^2}{b} = 1$

$\Rightarrow ax^2(a + b) – 2ax + (1 – b) = 0$

Or, $(-2a)^2 – 4a(a + b)(1 – b) = 0$

$\Rightarrow 4a^2 – 4a(a – ab + b – b^2) = 0$

$\Rightarrow ab + b^2 = b$ (as $a \neq 0$)

Hence, $a + b = 1$ (as $b \neq 0$)

**3.** If the difference of the roots of the equation $x^2 + px + q = 0$ be unity, prove that $p^2 + 4q^2 = (1 + 2q)^2.$

Solution 3:

Let $\alpha$ and $\beta$ be the roots of the given eqn.

Then, $\alpha + \beta = -p$ and $\alpha\beta = q.$

Now, by the question, $\alpha – \beta = 1$

$\Rightarrow (\alpha – \beta)^2 = (1)^2$

$\Rightarrow (\alpha + \beta)^2 – 4 \alpha \beta = 1$

Or, $p^2 – 4q = 1$

$\Rightarrow p^2 = 1 + 4q$

$\Rightarrow p^2 + 4q^2 = 1 + 4q + 4q^2$

$\therefore$ $p^2 + 4q^2 = (1 + 2q)^2.$

**4.** In writing a quadratic equation of the form $x^2 + bx + c = 0,$ a student writes the coefficient of x incorrectly and finds the roots as 7, 8. Another student writes the constant term incorrectly and finds the roots as 8, -3. Find the correct equation.

Solution 4:

Equation in condition 1:

$x^2 -(7 + 8)x + 7 xx 8 = 0$

We get, $c = 56$

Equation in condition 2:

$x^2 – (8 – 3)x + 8 xx (-3) = 0$

We get, $b = -5$

Hence, correct eqn. is:

$x^2 – 5x + 56 = 0.$

**5.** Show that $ax^2 + bx + c = 0$ and $x^2 + x + 1 = 0$ cannot have a common root unless $a = b = c.$

Solution 5:

Let’s consider $x^2 + x + 1 = 0$

$\Delta = 1^2 – 4.1.1 = -3 < 0$

Therefore, the roots are non-real complex and hence, if one root is common then both the roots must be common.

Now, non-real complex roots occur in conjugate pair. As such, the two equations are not distinct.

Hence, $\frac{a}{1} = \frac{b}{1} = \frac{c}{1}$

i.e. $a = b = c$

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