(i) (d)
(ii) (b)
(iii) (a): x(x + 3) + 7 = 5x - 11
\(\Rightarrow x^{2}+3 x+7=5 x-11\)
\(\Rightarrow x^{2}-2 x+18=0 \) is a quadratic equation.
\((b) (x-1)^{2}-9=(x-4)(x+3)\)
\(\Rightarrow x^{2}-2 x-8=x^{2}-x-12\)
\(\Rightarrow x-4=0\) is not a quadratic equation.
\((c) x^{2}(2 x+1)-4=5 x^{2}-10\)
\(\Rightarrow 2 x^{3}+x^{2}-4=5 x^{2}-10\)
\(\Rightarrow 2 x^{3}-4 x^{2}+6=0\) is not a quadratic equation.
\((d) x(x-1)(x+7)=x(6 x-9)\)
\(\Rightarrow x^{3}+6 x^{2}-7 x=6 x^{2}-9 x\)
\(\Rightarrow x^{3}+2 x=0\)   is not a quadratic equation.
(iv) (d)
(v) (d)

(i) (b):We have \(6 x^{2}+x-2=0\)
\(\Rightarrow \quad 6 x^{2}-3 x+4 x-2=0 \)
\(\Rightarrow \quad(3 x+2)(2 x-1)=0 \)
\(\Rightarrow \quad x=\frac{1}{2}, \frac{-2}{3}\)
(ii) (c): \(2 x^{2}+x-300=0\)
\(\Rightarrow \quad 2 x^{2}-24 x+25 x-300=0 \)
\(\Rightarrow \quad(x-12)(2 x+25)=0 \)
\(\Rightarrow \quad x=12, \frac{-25}{2}\)
(iii) (d): \(x^{2}-8 x+16=0\)
\(\Rightarrow(x-4)^{2}=0 \Rightarrow(x-4)(x-4)=0 \Rightarrow x=4,4\)
(iv) (d): \(6 x^{2}-13 x+5=0\)
\(\Rightarrow \quad 6 x^{2}-3 x-10 x+5=0 \)
\(\Rightarrow \quad(2 x-1)(3 x-5)=0 \)
\(\Rightarrow \quad x=\frac{1}{2}, \frac{5}{3}\)
(v) (a): \(100 x^{2}-20 x+1=0\)
\(\Rightarrow(10 x-1)^{2}=0 \Rightarrow x=\frac{1}{10}, \frac{1}{10}\)
 

(i) (b): Let two consecutive integers be x, x + 1.
Given, x² + (x + 1)² = 650
\(\begin{array}{l} \Rightarrow 2 x^{2}+2 x+1-650=0 \\ \Rightarrow 2 x^{2}+2 x-649=0 \end{array}\)
(ii) (c): Let the two numbers be x and 15 - x.
Given, \(\frac{1}{x}+\frac{1}{15-x}=\frac{3}{10}\)
\(\begin{array}{l} \Rightarrow 10(15-x+x)=3 x(15-x) \\ \Rightarrow 50=15 x-x^{2} \Rightarrow x^{2}-15 x+50=0 \end{array}\)
(iii) (d): Let the numbers be x and x + 3.
Given, x(x + 3) = 504
\(\Rightarrow\) x² + 3x - 504 = 0
(iv) (c): Let the number be x.
According to question, x² - 84 = 3(x + 8)
\(\Rightarrow x^{2}-84=3 x+24 \Rightarrow x^{2}-3 x-108=0\)
(v) (d): Let the number be x.
According to question, x + 12 = \(\frac {160}{x}\)
\(\Rightarrow x^{2}+12 x-160=0\)

(i) (b): Roots of the quadratic equation are 2 and -3.
\(\therefore\) The required quadratic equation is 
\((x-2)(x+3)^{n}=0 \Rightarrow x^{2}+x-6=0\)
(ii) (a): We have, 2x² + kx + 1 = 0
Since, -1/2 is the root of the equation, so it will satisfy the given equation
\(\therefore \quad 2\left(-\frac{1}{2}\right)^{2}+k\left(-\frac{1}{2}\right)+1=0 \Rightarrow 1-k+2=0 \Rightarrow k=3\)
(iii) (d): If the roots of the quadratic equations are opposites to each other, then coefficient of x (sum of roots) is 0.
So, both (a) and (b) have the coefficient of x = 0.
(iv) (c): The given equation is (x - 2)² + 19 = 0
\(\Rightarrow x^{2}-4 x+4+19=0 \Rightarrow x^{2}-4 x+23=0\)
(v) (b): If one root of a quadratic equation is irrational, then its other root is also irrational and also its conjugate i.e., if one root is p +.\(\sqrt(q)\) then its other root is p -.\(\sqrt(q)\).

(i) (a): To have no real roots, discriminant (D = b² - 4ac) should be < 0.
(a) D = 7² - 4(-4)(-4) = 49 - 64 = -15 < 0
(b) D=7²-4(-4)(-2)=49-32=17>0
(c) D = 5² - 4(-2)(-2) = 25 - 16 = 9 > 0
(d) D = 6² - 4(3)(2) = 36 - 24 = 12> 0
(ii) (b): To have rational roots, discriminant (D = b² - 4ac) should be> 0 and also a perfect square
(a) D = 1²- 4(1)( -1) = 1 + 4 = 5, which is not a perfect square.
(b) D = (-5)² - 4(1)(6) = 25 - 24 = I, which is a perfect square.
(c) D = (-3)² - 4(4)(-2) = 9 + 32 = 41, which is not a perfect square.
(d) D = (-1)² - 4(6)(11) = 1 - 264 = -263, which is not a perfect square.
(iii) (c) : To have irrational roots, discriminant (D = b² - 4ac) should be > 0 but not a perfect square.
(a) D = 2² - 4(3)(2) = 4 - 24 = -20 < 0
(b) D = (-7)² - 4(4)(3) = 49 - 48 = 1 > 0 and also a perfect square.
(c) D = (-3)² - 4(6)(-5) = 9 + 120 = 129> 0 and not a perfect square.
(d) D = 3² - 4(2)(-2) = 9 + 16 = 25 > 0 and also a perfect square.
(iv) (d): To have equal roots, discriminant (D = b² - 4ac) should be = 0.
(a) D=(-3)²-4(1)(4)=9-16=-7<0
(b) D = (-2)² - 4(2)(1) = 4 - 8 = -4 < 0
(c) D = (-10)² - 4(5)(1) = 100 - 20 = 80 > 0
(d) D = 6² - 4(9)(1) = 36 - 36 = 0
(v) (a): To have two distinct real roots, discriminant (D = b² - 4ac) should be > 0.
(a) D = 3² - 4(1)(1) = 9 - 4 = 5 > 0
(b) D = 3² - 4(-1)( -3) = 9 - 12 = -3 < 0
(c) D=8²- 4(4)(4) = 64-64 = 0
(d) D = 6² - 4(3)(4) = 36 - 48 = -12 < 0