Here a = 3, b = - 2 and \(c=\frac{1}{3} \text { . }\)
Therefore, discriminant b² - 4ac = (-2)² - 4 x 3 x \(\frac{1}{3}\) = 4 - 4 = 0.
Hence, the given quadratic equation has two equal real roots.
The roots are \(\frac{-b}{2 a}, \frac{-b}{2 a}, \text { i.e., } \frac{2}{6}, \frac{2}{6}, \text { i.e., } \frac{1}{3}, \frac{1}{3} .\)

(x+1)²=2(x-3)
x²+2x+1=2x-6
x²+2x-2x+1+6=0
x²+7=0
Which is of the form ax²+bx+c=0. Hence the given equation is a quadratic equation.

Let the quadratic polynomial be ax² + bx + c, and its zeroes be α and ß
Given \(\alpha+B=4=-\frac{b}{a}\) 
\(\alpha \beta=1=\frac{c}{a}\) 
If a = 1, then b = -4, c = 1
Therefore, the quadratic polynomial is x² – 4x +1.

Let the quadratic polynomial be ax² + bx + c, and its zeroes be α and ß
Given \(\alpha+\beta=-\frac{b}{a}=1\) 
\(\alpha \beta=\frac{c}{a}=1\) 
If a = 1, b = -1, and c = 1
Therefore, the quadratic polynomial is x² – x +1.

Let the polynomial be ax² + bx + c, and its zeroes be α and ß
Given \(\alpha+\beta=\sqrt{2}=\frac{-b}{a}\) 
\(\alpha \beta=\frac{1}{3}=\frac{c}{a}\) 
If a = 3, then \(b=\sqrt{2}\) and  c = 1/3
Therefore, the quadratic polynomial is \(3 x^{2}-3 \sqrt{2} x+1\).

We have
x² + 7x + 10 = (x + 2)(x + 5)
So, the value of x² + 7x + 10 is zero when x + 2 = 0 or x + 5 = 0, i.e., when x = – 2 or x = –5. Therefore, the zeroes of x² + 7x + 10 are – 2 and – 5. Now,
sum of zeroes = \(-2+(-5)=-(7)=\frac{-(7)}{1}=\frac{-(\text { Coefficient of } x)}{\text { Coefficient of } x^{2}}\) 
product of zeroes = \((-2) \times(-5)=10=\frac{10}{1}=\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}\)

(i) The number of zeroes is 1 as the graph intersects the x-axis at one point only.
(ii) The number of zeroes is 2 as the graph intersects the x-axis at two points.
(iii) The number of zeroes is 3. (Why?)
(iv) The number of zeroes is 1. (Why?)
(v) The number of zeroes is 1. (Why?)
(vi) The number of zeroes is 4. (Why?)

\(\sqrt { 2x^{ 2 } } +7x+5\sqrt { 2 } =0\)
\(\Rightarrow \sqrt { 2 } x^{ 2 }+2x+5x+5\sqrt { 2 } =0\)
\(\Rightarrow \sqrt { 2x } (x+\sqrt { 2 } )+5(x+\sqrt { 2 } )=0\)
\(\Rightarrow (x+\sqrt { 2 } )(\sqrt { 2x } +5)=0\)
\(\Rightarrow \) Either  \(x+\sqrt { 2 } =0\) or \(\sqrt { 2x } +5=0\)
\(\Rightarrow x=\sqrt { 2 } \) or \(x=-\frac { 5 }{ \sqrt { 2 } } \)

Here, the perimeter of a right triangle= 60 cm
Length of the hypotenuse = 25 cm
Let the base of the right triangle be x cm
\(\therefore \)  Perpendicular of the right triangle = 60 - 25 -x
= (35 - x) cm
 By using Pythagoras Theorem, we have 
x²+(35-x)²= (25)²
\(\Rightarrow x^{ 2 }+(35-x)^{ 2 }=(25)^{ 2 }\)
\(\Rightarrow x^{ 2 }+1225+x^{ 2 }-70x=625\)
\(\Rightarrow 2x^{ 2 }-70x+600=0\)
or x²-35x+300=0
x²-15x-20x+300=0
\(\Rightarrow x(x-15)-20x(x-15)=0\)
\(\Rightarrow (x-15)(x-20)=0\)
\(\Rightarrow x=15\) or x=20
When x = 15, Base = 15 cm, Altitude = 35 - 15 = 20 cm
When x = 20, Base = 20 cm Altitude 15 cm
Now, area of the right triangle 
\(=\frac { 1 }{ 2 } \)x Base x altitude
\(=\frac { 1 }{ 2 } \) x15x20 or \(\frac { 1 }{ 2 } \) x20x15
= 150 cm²

2

k < 4

29 years, 5 years

k = 14

\(-{\sqrt {2}\over 6} , \sqrt{2}\)

13 , 7

12 , 13

5 , 6

For coincident roots,D=0
\(\Rightarrow \) [-(2+m)]²-4X1(-m²-4m-4)=0
\(\Rightarrow \) (2+m)²+4(m²+4m+4)=0
\(\Rightarrow \) (2+m)²+4(m+2)²=0
\(\Rightarrow \) 5(2+m)²=0
\(\Rightarrow \) (2+m)²=0
\(\Rightarrow \) m=-2

Here a= \(\alpha -12,b=2(\alpha -12),c=2\)
For equal roots D=0 \(\Rightarrow \) b²-4ac=0
\(\Rightarrow \) [2( \(\alpha -12)]^{ 2 }\) -4X[2 \(\alpha -12)]=0\)
\(2(\alpha -12)[2\alpha -12)]=0\)
\(\Rightarrow \left( \alpha -12 \right) [(2\alpha -12)-4]=0\)
\(\Rightarrow \alpha =12,14\)
\(\alpha =12\) is not possible take \(\alpha =14\)

For equal roots, D = 0.

This is of the form ax²+bx+c=0, where a=4, b=4\(\sqrt { 3 } \) and c=3.
Discriminant(D)=b²-4ac=\({ \left( 4\sqrt { 3 } \right) }^{ 2 }\)-4 x 4 x 3=48-48=0
Roots are \(\alpha =\frac { -b+\sqrt { D } }{ 2a } =\frac { -4\sqrt { 3 } +0 }{ 8 } =\frac { -4\sqrt { 3 } }{ 8 } =\frac { -\sqrt { 3 } }{ 8 } \)
and \(\beta =\frac { -b-\sqrt { D } }{ 2a } =\frac { -4\sqrt { 3 } -0 }{ 8 } =\frac { -4\sqrt { 3 } }{ 8 } =\frac { -\sqrt { 3 } }{ 2 } \)
Hence, the roots are \(\frac { -\sqrt { 3 } }{ 2 } ,\frac { -\sqrt { 3 } }{ 2 } \)

Given, equation is \( \sqrt2x^2+7x+5\sqrt2=0\) 
\(\Rightarrow\) \( \sqrt2x^2+5x+2x+5\sqrt2=0\) [\(\because\) 5 \(\times\)2 = 10 and 5 + 2 = 7]
\(\Rightarrow\)\(x(\sqrt2x+5)+\sqrt2(\sqrt2x+5)=0\)
\(\Rightarrow\) (\(\sqrt{2}\)x + 5) (x + \(\sqrt{2}\)) = 0
\(\Rightarrow\) \(\sqrt{2}\)x+5 = 0 or x + \(\sqrt{2}\) = 0
\(\Rightarrow\) x=\(\frac{-5}{\sqrt{2}}\) or x = -\(\sqrt{2}\)
Hence, the roots of the equation
\(\sqrt{2} x^2+7 x+5 \sqrt{2}=0 \text { are } \frac{-5}{\sqrt{2}} \text { and }-\sqrt{2} \text {. }\)

Let present age of Rohan be x years.
Rohan's mother's present age be (x+26+3)years.
After 3 years, Rohan's age = (x+3) years.
After 3 years, Rohan's mother's age= (x+26+3)years
ATQ (x+3)(x+29)=360
x²+32x-273=0
Which is the required quadratic equation.

Let the two consecutive positive integers be x and x+1
Then, according to the question,
x(x+1) = 306 \(\Rightarrow\) x² + x = 306
\(\Rightarrow\) x² + x - 306 = 0,
Which is the required quadratic equation.

Given, equation is (x+2)³=2x(x²-1)
\(\Rightarrow\) x³ + 8 + 3x²(2) + 3x (2)² = 2x³ - 2x
[\(\because\) (a + b)³ = a³ + b³ + 3a²b + 3ab²]
\(\Rightarrow\) x³ + 8 + 6x² + 12x = 2x³ - 2x
\(\Rightarrow\) x³ + 8 + 6x² + 12x - 2x³ + 2x = 0
\(\Rightarrow\) -x³ + 6x² + 14x + 8 = 0
which is not of the form ax² + bx + c = 0,
because it has cubic term, i.e. x³.
\(\therefore\) It is not a quadratic equation.