(3)

$D=b^2–4ac=4^2–4\times 2\times 3=16–24=–8<0$

Since $D<0$

Hence, roots are not real.

(2)

Factors of $p=p\times 1$

$\therefore$ $\text{Roots are }p\text{ and }1.$

The quadratic equation is:

$x^2-\text{(sum of roots)}x+\text{product of roots}=0$

$\Rightarrow x^2-(p+1)x+p=0$

(3)

If the discriminant is equal to zero, i.e., $b^2-4ac=0$ where a, b, c are real numbers and $a\ne 0,$ then roots of the quadratic equation $a{x}^2+bx+c=0,$ are real and equal $b^2-4ac=0\Rightarrow ac=\frac{b^2}{4}$

(2)        – 11

(4)

Given equation is $x^2+x+1=0$

Where $a=1,$ $b=1,$ $c=1$

$D=b^2-4ac={\left(1\right)}^2-4\times 1\times 1$

$\Rightarrow D=-3$

Where $D<0$

When $D<0$ roots are not-real.

(1)

Given equation is $x^{2/3} + x^{1/3}- 2 = 0$

⇒ t² + t - 2 = 0     (Take $x^{1/3}$ = t)

⇒ t² + 2t - t - 2 = 0

⇒ t(t + 2) - 1(t + 2) = 0     ⇒ (t + 2)(t - 1) = 0

The roots of above equation are t = 1 and t = -2.

If t = 1, then x = 1.

If t = -2, then x = -8.

Since p is a root of the quadratic equation

$x^2-(p+q)x+k=0,$
we have
$p^2-(p+q)p+k=0\Rightarrow p^2-p^2-pq+k=0\Rightarrow k=pq$

(4)

$$\begin{gathered} Given quadratic equation: 4x² – \surd 3x – 5 = 0. \\ Sum of roots = -b/a = \surd 3/4 \end{gathered}$$

(1)

$$\begin{gathered} Given equation: x² + kx – 5/4 = 0. Since ½ is a root, \\ (½)² + k ½ – 5/4 = 0 \Rightarrow 1/4 + k/2 – 5/4 = 0 \Rightarrow k/2 – 1 = 0 \Rightarrow k = 2 \end{gathered}$$

(2)

Given quadratic equation: $9x² + 3/4 x – \surd 2 = 0$.

Product of roots =$c/a = -\surd 2/9$

(2)

Given quadratic equation: $5x² – 13x + k = 0$. Since one root is reciprocal of the other,

Product of roots $= 1 \Rightarrow c/a = 1 \Rightarrow c = a \Rightarrow k = 5$

(4)

$Since 2 is a root, 2\left(2\right)² – 10\left(2\right) + p = 0 \Rightarrow 8 – 20 + p = 0 \Rightarrow p = 12$

(2)

As observed, the sum of roots is –5 and product of roots is –(a + 1)(a + 6). This is possible only when the roots are (a + 1) and –(a + 6).

(3)

Given, x = –2 and 3/4 are roots of the quadratic equation $Ax² + Bx – 6 = 0.$

Product of roots =$–6/A \Rightarrow (–2 \times 3/4) = –6/A \Rightarrow A = 4$
Sum of roots = $–B/A \Rightarrow (–2 + 3/4) = –B/4 \Rightarrow –5/4 = –B/4 \Rightarrow B = 5$

Both A and B are positive numbers and sum of A and B is an odd number.

(3)

Given equation is $ax² + bx + c = 0, a \ne 0.$
For real and equal roots, D = 0.

⇒$b² - 4ac = 0 \Rightarrow b² = 4ac \Rightarrow b²/4 = ac$

(4)

Given quadratic equation $ax² + bx + c = 0, a \ne 0.$
For real and equal roots, discriminant = 0.

⇒$b² - 4ac = 0 \Rightarrow b² = 4ac \Rightarrow c = b²/4a$

(4)

$Given x² - 4x + k = 0. For distinct real roots, D > 0.$

$\Rightarrow (-4)² - 4\times 1\times k > 0 \Rightarrow 16 - 4k > 0 \Rightarrow 16 > 4k \Rightarrow k < 4.$

(2)

Given quadratic equation is $2x² + kx + 2 = 0.$
For equal roots, discriminant = 0.

$\Rightarrow k² - 4\times 2\times 2 = 0 \Rightarrow k² - 16 = 0 \Rightarrow k = \pm 4$

(3)

Given equation: $(x² + 1)² - x² = 0 \Rightarrow (x² + 1 - x)(x² + 1 + x) = 0.$

$\Rightarrow (x² - x + 1)(x² + x + 1) = 0 or x² - x + 1 = 0,$
$Its discriminant D = (-1)² - 4\times 1\times 1 = -3 < 0$
$Its discriminant D = 1² - 4\times 1\times 1 = -3 < 0$

So, there are no real roots.

(2)

For real and distinct roots, the discriminant (D) of the equation 9$x² - 3kx + k = 0$must be greater than zero.

$D = (-3k)² - 4\times 9\times k > 0 \Rightarrow 9k² - 36k > 0 \Rightarrow 9k(k - 4) > 0$

This inequality implies that k < 0 or k > 4.
Therefore, the possible values of k are (iii) and (iv).

(1)

$D₁ = a² - 4b and D₂ = c² - 4d$

$Now, D₁ + D₂ = a² + c² - 4(b + d) = a² + c² - 2ac [Since 2(b + d) = ac]$

$\Rightarrow D₁ + D₂ = (a - c)² \ge 0$

So, at least one of $D₁ and D₂$ is greater than or equal to zero. Thus, at least one of the given two equations has real roots.