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#### If b = 0, c < 0, is it true that the roots of x2 + bx+ c = 0 are numerically equal and opposite in sign? Justify.

True

Quadratic Equation – A quadratic equation in x is an equation that can be written in the standard form, ax2 + bx + c = 0 Where a, b and c are real numbers with a $\neq$ 0

Roots : If ax2 + bx + c = 0                  …..(1)

Is a quadratic equation then the values of x which satisfy equation 1 are the roots of the equation.

Here the given equation is x2 + bx+ c = 0      …..(2)

It is also given that b = 0, c < 0.

Let       c=-y

Put       b = 0, c = – y  in (2)
$\\x^{2}+0(x)-y=0\\ x^{2}=y\\ x=\pm \sqrt{y}\\ x=+\sqrt{y}\; \; \; \; \; \; \; \; \; \; x=-\sqrt{y}$

Hence both the roots are equal and opposite in sign. Hence the given statement is true.

#### Is 0.2 a root of the equation x2 – 0.4 = 0?  Justify

False

Quadratic Equation – A quadratic equation in x is an equation that can be written in the standard form, ax2 + bx + c = 0    Where a, b and c are real numbers with a $\neq$ 0

Roots : If ax2 + bx + c = 0                  …..(1)

Is a quadratic equation then the values of x which satisfy equation 1 are the roots of the equation.

Here the given equation is x2 – 0.4 = 0          …..(2)

If 0.2 is a root of equation 2 then it should satisfy its equation
Put x = 0.2 in (2)

$\\(0.2)^{2}-0.4=0\\ \left (\frac{2}{10} \right )-0.4=0\\ \frac{4}{100}-0.4=0\\ 0.04-0.4=0\\ -0.36 \neq 0$

Here 0.2 is not satisfying the equation (2)

Hence 0.2 is not a root of the equation $x^{2}-0.4=0$

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#### Does there exist a quadratic equation whose coefficients are all distinct irrationals but both the roots are rational? Why?

True

Let us consider an equation with distinct irrational numbers

$\sqrt{2}x^{2}-5\sqrt{2}x+4\sqrt{2}=0$

compare with $ax^{2} + bx + c = 0$ where $a \neq 0$

$a=\sqrt{2},b=-5\sqrt{2},c=4\sqrt{2}$

let us find the roots of the equation

$\\x=\frac{-b+\sqrt{b^{2}-4ac}}{2a}\\ x=\frac{5\sqrt{2 }\pm \sqrt{50-4(\sqrt{2})(4\sqrt{2})}}{2 \sqrt{2}}\\ x=\frac{5 \sqrt{2} \pm \sqrt{50-32}}{2\sqrt{2}}\\ x=\frac{5 \sqrt{2} \pm \sqrt{18}}{2\sqrt{2}}\\ x=\frac{\sqrt{2}(5\pm 3)}{2\sqrt{2}}=\frac{5 \pm 3}{2}\\ x=\frac{5+3}{2}=4 \; \; \; \; \; \; \; \; x=\frac{5-3}{2}=1$

The roots are rational. Hence the given statement is true

#### Does there exist a quadratic equation whose coefficients are rational but both of its roots are irrational? Justify your answer.

True

Solution

Rational number – A number that can be expressed in the form of $\frac{p}{q}$  where $q\neq 0$

(p, q are integers)

Irrational number – A number that can not be expressed in the form of ratio of two integers.

Let us suppose a quadratic equation with rational coefficient.

$-x^{2}-2x+4=0$

compare with $ax^{2} + bx + c = 0$ where $a \neq 0$

Let us find the roots of the equation

$x=\frac{-b+\sqrt{b^{2}-4ac}}{2a}\\ x=\frac{2 \pm \sqrt{4-4(-1)(4)}}{2 (-1)}\\ x=\frac{2 \pm \sqrt{20}}{-2}\\ x=\frac{2\pm2\sqrt{5}}{-2}\\ x=\frac{2+2\sqrt{5}}{-2} \; \; \; \; \; \; \; \; x=\frac{2-2\sqrt{5}}{-2}\\$

Both of its roots are irrational.

Hence the given statement is true.

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False

Solution

Quadratic Equation – A quadratic equation in x is an equation that can be written in the standard form, $ax^{2} + bx + c = 0$

Where a, b and c are real numbers with $a \neq 0$

Root: If $ax^{2} + bx + c = 0$                     …..(1)

Is a quadratic equation then the values of x which satisfy equation (1) are the roots of the equation.

Let us take a quadratic equation with integral coefficient

$2x^{2}-3x-5=0$

compare with $ax^{2} + bx + c = 0$ where $a \neq 0$

(a=2,b=-3,c=-5)

let us find the roots of the equation

$\\x=\frac{-b+\sqrt{b^{2}-4ac}}{2a}\\ x=\frac{3 \pm \sqrt{9-4(2)(-5)}}{2 \times 2}\\ x=\frac{3 \pm \sqrt{9+40}}{4}\\ x=\frac{3+\sqrt{49}}{4}\\ x=\frac{3+7}{4}=\frac{5}{7} \; \; \; \; \; \; \; \; x=\frac{3-7}{4}=-1$

Here we found that 5/2 is not on integral

Hence the given statement is false

#### Write whether the following statements are true or false. Justify your answers.(i) Every quadratic equation has exactly one root.(ii) Every quadratic equation has at least one real root.(iii) Every quadratic equation has at least two roots.(iv) Every quadratic equations has at most two roots.(v) If the coefficient of x2 and the constant term of a quadratic equation have opposite signs, then the quadratic equation has real roots.(vi)If the coefficient of x2 and the constant term have the same sign and if the coefficient of x term is zero, then the quadratic equation has no real roots.

(i) False

Solution

Quadratic Equation – A quadratic equation in x is an equation that can be written in the standard form,             $ax^{2} + bx + c = 0$

Where a, b and c are real numbers with $a \neq 0$

Root: If $ax^{2} + bx + c = 0$                     …..(1)

Is an quadratic equation then the values of x which satisfy equation (1) are the roots of the equation.

Let us consider a quadratic equation  $x^{2}-4=0$
$x^{2}-4=0\\ x^{2}=4\\ x=\pm \sqrt{4}\\ x=2 \: \: \: \: \: \: \: x=-2$

Here – 2, 2 are the two roots of the equation.

Hence it is false that every quadratic equation has exactly one root.

(ii) False

Solution

Quadratic Equation – A quadratic equation in x is an equation that can be written in the standard form,             $ax^{2} + bx + c = 0$

Where a, b and c are real numbers with $a \neq 0$

Root: If $ax^{2} + bx + c = 0$                     …..(1)

Is an quadratic equation then the values of x which satisfy equation (1) are the roots of the equation.

Let us consider a quadratic equation  $x^{2}-x+2=0$
$x^{2}-x+2=0$

compare with $ax^{2} + bx + c = 0$ where $a \neq 0$

$a=1,b=-1,c=2\\ b^{2}-4ac=(-1)^{2}-4(1)(2)\\ =1-8\\ =-7\\ b^{2}-4ac<0$

Hence both the roots of the equation are imaginary.

Hence the statement every quadratic equation has at least one real root is False.

(iii) False

Solution

Quadratic Equation – A quadratic equation in x is an equation that can be written in the standard form,             $ax^{2} + bx + c = 0$

Where a, b and c are real numbers with $a \neq 0$

Root: If $ax^{2} + bx + c = 0$                     …..(1)

Is an quadratic equation then the values of x which satisfy equation (1) are the roots of the equation.

Let us consider a quadratic equation  $x^{2}-4x+4=0$

$x^{2}-4x+4=0\\ x^{2}+(2)^{2}-2(x)(2)\\ (x-2)^{2}=0\: \: \: \: \: \: \: \: \: \: \: \: \: \: (using\: (a-b)^{2}=a^{2}+b^{2}-2ab)\\ (x-2)=0\\ x=2$

The equation $x^{2}-4x+4=0$  has only one root which is x = 2

Hence the given statement is False

(iv) True

Quadratic Equation – A quadratic equation in x is an equation that can be written in the standard form,             $ax^{2} + bx + c = 0$

Where a, b and c are real numbers with $a \neq 0$

Root: If $ax^{2} + bx + c = 0$                     …..(1)

Is an quadratic equation then the values of x which satisfy equation (1) are the roots of the equation.

As we know that the standard form of a quadratic equation is $ax^{2} + bx + c = 0$

It is a polynomial of degree 2.

As per the power of x is 2. There is at most 2 values of x exist that satisfy equation.

Hence the given statement every quadratic equations has at most two roots is true.

(v) True

Quadratic Equation – A quadratic equation in x is an equation that can be written in the standard form,             $ax^{2} + bx + c = 0$

Where a, b and c are real numbers with $a \neq 0$

Root: If $ax^{2} + bx + c = 0$                     …..(1)

Is an quadratic equation then the values of x which satisfy equation (1) are the roots of the equation.

Let us suppose a quadratic equation $x^{2} + x - 2 = 0$

$x^{2} + x - 2 = 0$  is a quadratic equation in which coefficient of x2 and constant term have opposite signs.

$x^{2} + x - 2 = 0$

compare with $ax^{2} + bx + c = 0$ where  $a \neq 0$
$a=1,b=1,c=-2\\ b^{2}-4ac=(1)^{2}-4(1)(-2)\\ =1+8\\ =9\\ b^{2}-4ac>0$

Here $b^{2}-4ac>0$

Hence these types of equations have real roots.

(vi) True

Quadratic Equation – A quadratic equation in x is an equation that can be written in the standard form,             $ax^{2} + bx + c = 0$

Where a, b and c are real numbers with $a \neq 0$

Root: If $ax^{2} + bx + c = 0$                     …..(1)

Is an quadratic equation then the values of x which satisfy equation (1) are the roots of the equation.

Let us suppose a quadratic equation $x^{2} + 4= 0$

In $x^{2} + 4= 0$, the coefficient of x2 and constant term has same sign and coefficient of x is 0.

$x^{2} + 4= 0$

Compare with $ax^{2} + bx + c = 0$ where $a \neq 0$
$a=1,b=0,c=4\\ b^{2}-4ac=(0)^{2}-4(1)(4)\\ =16\\ b^{2}-4ac<0$

Here $b^{2}-4ac<0$

Hence these types of equations have no real roots.

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#### State whether the following quadratic equations have two distinct real roots. Justify your answer.$(i)x^{2}-3x+4=0$         $(ii)2x^{2}+x-1=0$$(iii)2x^{2}-6x+\frac{9}{2}=0$$(iv)3x^{2}-4x+1=0$$(v)(x+4)^{2}-8x=0$$(vi)(x-\sqrt{2})^{2}-2(x+1)=0$$(vii)\sqrt{2}x^{2}-\frac{3}{\sqrt{2}}x+\frac{1}{2}=0$$(viii)x(1-x)-2=0$$(ix)(x-1)(x+2)+2=0$$(x)(x+1)(x-2)+x=0$

(i)     Quadratic equation : A quadratic equation in x is an equation that can be written in the standard form $ax^{2}+bx+c=0$  Where a, b and c are real numbers with $a\neq 0$

The given equation is $x^{2}-3x+4=0$

Here a=1,b=-3,c=4
$D=b^{2}-4ac\\ =(-3)^{2}-4(1)(4)\\ =9-16=-7\\ b^{2}-4ac<0$

Hence the equation has no real roots.

(ii)   Quadratic equation : A quadratic equation in x is an equation that can be written in the standard form $ax^{2}+bx+c=0$           Where a, b and c are real numbers with $a\neq 0$

The given equation is $2x^{2}+x-1=0$

Here a=2,b=-1,c=-1
$D=b^{2}-4ac\\ =(1)^{2}-4(2)(-1)\\ =1+8=9\\ b^{2}-4ac>0$

Here for the equation $2x^{2}+x-1=0$$b^{2}-4ac>0$

Hence it is a quadratic equation with two distinct real roots

(iii)   Quadratic equation : A quadratic equation in x is an equation that can be written in the standard form $ax^{2}+bx+c=0$           Where a, b and c are real numbers with $a\neq 0$

The given equation is $2x^{2}-6x+\frac{9}{2}=0$

Here $a= 2,b=-6,c=\frac{9}{2}$

$D=b^{2}-4ac\\ =(-6)^{2}-4(2)(\frac{9}{4})\\ =36-36=0\\ b^{2}-4ac=0$

here $b^{2}-4ac=0$

Hence the equation have two real and equal roots.

(iv)    Quadratic equation : A quadratic equation in x is an equation that can be written in the standard form $ax^{2}+bx+c=0$           Where a, b and c are real numbers with $a\neq 0$

The given equation is $3x^{2}-4x+1=0$

Here a=3,b=-4,c=1
$D=b^{2}-4ac\\ =(-4)^{2}-4(3)(1)\\ =16-12=4\\ b^{2}-4ac>0$

Here  $b^{2}-4ac>0$

Hence the equation has distinct and real roots.

(v) Quadratic equation : A quadratic equation in x is an equation that can be written in the standard form $ax^{2}+bx+c=0$           Where a, b and c are real numbers with $a\neq 0$

The given equation is $(x+4)^{2}-8x=0$

$(x)^{2}+(4)^{2}+2(x)(4)-8x=0$                                              $\left (using (a+b)^{2}=(a)^{2}+(b)^{2}+2ab \right )$

$(x)^{2}+16+8x-8x=0\\ x^{2}+16=0$

$a=1,b=0,c=16\\ =(0)^{2}-4(1)(16)\\ =-64\\ b^{2}-4ac<0$

Here  $b^{2}-4ac<0$

Hence the equation has no real roots.

(vi) Quadratic equation : A quadratic equation in x is an equation that can be written in the standard form $ax^{2}+bx+c=0$           Where a, b and c are real numbers with $a\neq 0$

The given equation is $(x-\sqrt{2})^{2}-2(x+1)=0$

$(x)^{2}+(\sqrt{2})^{2}+2(x)(\sqrt{2})-2x-2=0$                        $\left (using (a+b)^{2}=(a)^{2}+(b)^{2}+2ab \right )$

$(x)^{2}+2-2\sqrt{2}x-2x-2=0\\ x^{2}-(2\sqrt{2}+2)x=0$

Compare with $ax^{2}+bx+c=0$ where  $a\neq 0$

$a=1,b=-(2\sqrt{2}+2),c=0\\b^{2}-4ac =(-(2\sqrt{2}+2))^{2}-4(1)(0)\\ =(2\sqrt{2}+2)^{2}-0$

$=(2\sqrt{2})^{2}+(2)^{2}+2(2\sqrt{2})(2)$                            $\left (using (a+b)^{2}=(a)^{2}+(b)^{2}+2ab \right )$

$=8+4+8\sqrt{2}\\ =12+8\sqrt{2}\\$

$b^{2}-4ac>0$

Here  $b^{2}-4ac>0$

Hence the equation has two distinct real roots.

(vii)    Quadratic equation : A quadratic equation in x is an equation that can be written in the standard form $ax^{2}+bx+c=0$           Where a, b and c are real numbers with $a\neq 0$

The given equation is $\sqrt{2}x^{2}-\frac{3}{\sqrt{2}}x+\frac{1}{2}=0$

Here $a=\sqrt{2},b=\frac{3}{\sqrt{2}},c=\frac{1}{2}$
$b^{2}-4ac =\left (- \frac{3}{\sqrt{2}} \right )^{2}-4(\sqrt{2})\left (\frac{1}{2} \right )\\ =\frac{9}{2}-2\sqrt{2}\\=4.5-3.8=1.7\\ b^{2}-4ac>0$

Here  $b^{2}-4ac>0$

Hence the equation has distinct real roots.

(viii) Quadratic equation : A quadratic equation in x is an equation that can be written in the standard form $ax^{2}+bx+c=0$           Where a, b and c are real numbers with $a\neq 0$

The given equation is $x(1-x)-2=0$

$x-(x)^{2}-2=0\\ -x^{2}+x-2=0$

compare with $ax^{2}+bx+c=0$ where $a\neq 0$

$a=-1,b=1,c=-2\\ b^{2}-4ac=(1)^{2}-4(-1)(-2)\\1-8=-7\\ b^{2}-4ac<0$

Here  $b^{2}-4ac<0$

Hence the equation has no real roots.

(ix) Quadratic equation : A quadratic equation in x is an equation that can be written in the standard form $ax^{2}+bx+c=0$           Where a, b and c are real numbers with $a\neq 0$

The given equation is $(x-1)(x+2)+2=0$

$x(x+2)-1(x+2)+2=0\\ x^{2}+2x-x-2+2=0\\ x^{2}+x=0\\$

$a=1,b=1,c=0\\b^{2}-4ac =(1)^{2}-4(1)(0)\\ =1$

$b^{2}-4ac>0$

Here  $b^{2}-4ac>0$

Hence the equation has two distinct real roots.

(x) Quadratic equation : A quadratic equation in x is an equation that can be written in the standard form $ax^{2}+bx+c=0$           Where a, b and c are real numbers with $a\neq 0$

The given equation is $(x+1)(x-2)+x=0$

$x(x-2)+1(x-2)+x=0\\ x^{2}-2x+x-2+x=0\\ x^{2}-2x+2x-2=0\\x^{2}-2=0$

Compare with $ax^{2}+bx+c=0$ where $a\neq 0$

$a=1,b=0,c=-2\\b^{2}-4ac =(0)^{2}-4(1)(-2)\\ =8$

$b^{2}-4ac>0$

Here  $b^{2}-4ac>0$

Hence the equation has two distinct real roots.

#### Write whether the following statement is True or False. Justify your answer.The graph of every linear equation in two variables need not be a line.

False

Solution:

We know that the standard form of any linear equation is ax + by + c = 0

If we put different value of x we get different values of y corresponding to the values of x.

So we have x proportional to y.

Thus we can say $\frac{x}{y}$ = constant.

Now, the general equation of a line is $y = mx + c$, which also gives a direct proportionality between x and y.

So if we plot a graph with the help of such points, it will always be a line

Therefore the given statement is false.

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#### Write whether the following statement is True or False. Justify your answer.Every point on the graph of a linear equation in two variables does not represent a solution of the linear equation.

False

Solution:

The graph of a linear equation in two variables is a line which can be plotted by finding two solutions (a1, b1) and (a2, b2).

Hence the graph of a linear equation is constructed by joining these points. So any point on the graph must be a solution of this linear equation.

Hence the given statement is False.

#### Write whether the following statement is True or False. Justify your answer.The coordinates of points in the table: x 0 1 2 3 4 y 2 3 4 -5 6 represent some of the solutions of the equation x – y + 2 = 0.

False

Solution:

The given equation is x – y + 2 = 0

Put x = 0 in the given equation

0 – y + 2 = 0

y = 2, i.e., the point is (0, 2)

Put x = 1 in the given equation

1 – y + 2 = 0

y = 3, i.e., the point is (1, 3)

Put x = 2 in the given equation

2 – y + 2 = 0

4 = y, i.e., the point is (2, 4)

Put x = 3 in the given equation

3 – y + 2 = 0

5 = y, i.e., the point is (3, 5)

Put x = 4 in the given equation

4 – y + 2 = 0

6 = y, i.e., the point is (4, 6)

The coordinates of points in the table:

 x 0 1 2 3 4 y 2 3 4 -5 6

It is given that the above represents some of the solutions of the equation x – y + 2 = 0.

Here all the points satisfy the given equation except (3, -5).

Therefore the given statement is false because one of the table entry is an incorrect solution for the given equation.