Question 12 Marks
In each of the following, determine whether the given values are solution of the given equation or not:
$x^2 + x + 1 = 0; x = 0; x = 1$
AnswerNow Substitute $x=0$ in given equation
$\text { L.H.S. }=(0)^2+0+1 \neq 0 \neq \text { R.H.S. }$
on substituting $x=1$ in L.H.S. of given equation
$\Rightarrow(1)^2+1+1 \neq 0 \neq \text { R.H.S. }$
Hence $x=0$ and $x=1$ are not solutions of the given equation.
View full question & answer→Question 22 Marks
In each of the following determine whether the given values are solutions of the equation or not.
$x^2+\sqrt{2}-4=0 ; x=\sqrt{2}, x=-2 \sqrt{2}$
AnswerGiven equation
$x^2+\sqrt{2}-4=0 ; x=\sqrt{2}, x=-2 \sqrt{2}$
Substitute $x=\sqrt{2}$ in the L.H.S.
$\text { L.H.S. }=(\sqrt{2})^2+\sqrt{2} \times \sqrt{2}-4 $
$=2+2-4 $
$=4-4$
$=0$
Hence $x=\sqrt{2}$ is a solution of the given equation.
View full question & answer→Question 32 Marks
In each of the following determine whether the given values are solutions of the equation or not
$2 x^2-6 x+3=0 ; x=\frac{1}{2}$
AnswerGiven equation is
$2 x^2-6 x+3=0 ; x=\frac{1}{2}$
Substitute $x=\frac{1}{2}$ in L.H.S.
L.H.S. $=2 \times\left(\frac{1}{2}\right)^2-6 \times \frac{1}{2}+3$
$=2 \times \frac{1}{4}-3+3=\frac{1}{2} \neq 0$
Hence, $x=\frac{1}{2}$ is not a solution ofthe given equation.
View full question & answer→Question 42 Marks
In each of the following determine whether the given values are solutions of the equation or not.
$3x^2 - 2x - 1 = 0; x = 1$
AnswerGiven equation is
$3x^2 - 2x - 1 = 0; x = 1$
Put $x = 1$ in the $L.H.S.$
$L.H.S. = 3(1)^2 - 2 x 1 - 1$
$= 3 - 3$
$= 0$
$= R.H.S.$
Hence, $x = 1$ is a solution of the given equation.
View full question & answer→Question 52 Marks
Find the values of k so that the sum of tire roots of the quadratic equation is equal to the product of the roots in each of the following:
$kx^2 + 2x + 3k = 0$
Answer$k x^2+2 x+3 k=0$
Here, $a = k, b = 2,$ and $c = 3k.$
$\text { Sum of roots }=-\frac{b}{a}=-\frac{2}{k} $
$\text { Product of root }=\frac{c}{a}$
$=\frac{3 k}{k} $
$ =3$
Sum of roots = Product of roots
$-\frac{2}{k}=3 $
$ \Rightarrow 3 k =-2 $
$ \Rightarrow k =-\frac{2}{3} .$
View full question & answer→Question 62 Marks
Determine whether the given quadratic equations have equal roots and if so, find the roots:
$3x^2 - 6x + 5 = 0$
AnswerThe given equation is
$3x^2 - 6x + 5 = 0$
Here, $a = 3, b = -6$ and $c = 5$
Discriminant
$= b^2 - 4ac$
$= (-6)^2 - 4 x 3 x 5$
$= 36 - 60$
$= -24 < 0$
imaginary roots.
View full question & answer→Question 72 Marks
Solve the following quadratic equation by factorisation:
$2x^2 + ax - a^2 = 0$ where $a ∈ R.$
Answer$2 x^2+a x-a^2=0$
$ \Rightarrow 2 x^2+2 a x-a x-a^2=0$
$ \Rightarrow 2 x(x+a)-a(x+a)=0 $
$ \Rightarrow(x=a)(2 x-a)=0 $
$ \Rightarrow x+a=0 \text { or } 2 x-a=0$
$ \Rightarrow x=-a \text { and } x=\frac{a}{2} .$
View full question & answer→Question 82 Marks
Solve the following quadratic equation by factorisation:
$9x^2 - 3x - 2 = 0$
AnswerThe given quadratic equation is
$9 x^2-3 x-2=0 \\ \Rightarrow 9 x^2-6 x+3 x-2=0$
$ \Rightarrow 3 x(3 x-2)+1(3 x-2)=0$
$\Rightarrow(3 x-2)(3 x+1)=0 $
$ \Rightarrow 3 x-2=0 \text { or } 3 x+1=0$
$ \Rightarrow 3 x=2 \text { and } 3 x=-1 $
$ \Rightarrow x=\frac{2}{3} \text { and } x=-\frac{1}{3} .$
View full question & answer→Question 92 Marks
Solve the following quadratic equation by factorisation:
$x^2 - 3x - 10 = 0$
Answer$x^2 - 3x - 10 = 0$
$\Rightarrow x^2- 5x + 2x - 10 = 0$
$\Rightarrow x(x - 5) + 2(x - 5) = 0$
$\Rightarrow (x - 5) (x + 2) = 0$
$\Rightarrow x - 5 = 0$ or $x + 2 = 0$
$⇒ x = 5$ and $x = -2$
View full question & answer→Question 102 Marks
Solve the following quadratic equation by factorisation:
$(2x + 3) (3x - 7) = 0$
AnswerThe given quadratic equation is
$(2 x+3)(3 x-7)=0$
$ \Rightarrow 2 x+3=0 \text { or } 3 x-7=0 $
$\Rightarrow x=\left\{\frac{-3}{2}, \frac{7}{3}\right\} .$
View full question & answer→Question 112 Marks
Solve the following quadratic equation by factorisation:
(x - 4) (x + 2) = 0
AnswerThe given quadratic equation is
(x - 4) (x + 2) = 0
⇒ x - 4 = 0 or x + 2 = 0
⇒ x = {4, -2}.
View full question & answer→Question 122 Marks
In each of the following find the values of k of which the given value is a solution of the given equation:
$x^2 + 3ax + k = 0; x = a.$
Answer$x^2 + 3ax + k = 0; x = a.$
Putting $x = -a$ in $L.H.S$. of equation
$L.H.S = (-a)^2 + 3a x (-a) + k = 0$
$\Rightarrow a^2- 3a^2 + k = 0$
$\Rightarrow -2a^2 + k = 0$
Hence, $k = 2a^2.$
View full question & answer→Question 132 Marks
In each of the following find the values of k of which the given value is a solution of the given equation:
$k x^2+\sqrt{2} x-4=0 ; x=\sqrt{2}$
Answer$k x^2+\sqrt{2} x-4=0 ; x=\sqrt{2}$
Putting $x=\sqrt{ } 2$ in L.H.S. of equation
$\text { L.H.S }= k \cdot(\sqrt{2})^2+\sqrt{2} \times \sqrt{2}-4=0$
$\Rightarrow 2 k +2-4=0$
$\Rightarrow 2 k -2=0$
Hence, $k =\frac{2}{2}$.
View full question & answer→Question 142 Marks
In each of the following find the values of k of which the given value is a solution of the given equation:
$x^2 - x(a + b) + k = 0, x = a$
Answer$x^2 - x(a + b) + k = 0, x = a.$
Putting $x = a$ in $L.H.S$. of equation
$\Rightarrow (a)^2 - a(a + b) + k = 0$
$\Rightarrow a^2 - a^2- ab + k = 0$
Hence,$ k = ab.$
View full question & answer→Question 152 Marks
In each of the following find the values of k of which the given value is a solution of the given equation:
$7 x^2+k x-3=0 ; x=\frac{2}{3}$
Answer$7 x ^2+ kx -3=0 ; x =\frac{2}{3} $
$ \text { Putting } x =\frac{2}{3} \text { in L.H.S. of equation } $
$ \text { L.H.S. }=7 \times\left(\frac{2}{3}\right)^2+\frac{2}{3} k -3=0 $
$ \Rightarrow \frac{28}{9}+\frac{2}{3} k -3=0 $
$ \Rightarrow \frac{28+6 k -27}{9}=0 $
$ \Rightarrow 6 k +1=0$
Hence, $k=-\frac{1}{6}$
View full question & answer→Question 162 Marks
$48x^2 – 13x -1 = 0$
Answer$48 x^2-13 x-1=0 $
$\Rightarrow 48 x^2-16 x+3 x-1=0 $
$ \Rightarrow 16 x(3 x-1)+1(3 x-1)=0 $
$ \Rightarrow(3 x-1)(16 x+1)=0$
$ \Rightarrow 3 x-1=0 \text { or } 16 x+1=0$
$x=\frac{1}{3}$ or $x=\frac{-1}{16}$ are two roots of the equation.
View full question & answer→Question 172 Marks
An aeroplane travelled a distance of 400 km at an average speed of x km/hr. On the return journey the speed was increased by 40 km/hr. Write down the expression for the time taken for
The outward journey
AnswerTime taken for the onward journey
$=\frac{400}{x}$ hours.
View full question & answer→Question 182 Marks
Answer(3x - 5)(2x + 7) = 0
2x + 7 = 0 or 3x - 5 = 0
$x =-\frac{7}{2}$ or $x=\frac{5}{3}$
Hence x $=\frac{5}{3}$ and $x=-\frac{7}{2}$ are two roots of the equation.
View full question & answer→Question 192 Marks
Find if x = – 1 is a root of the equation $2x^2 – 3x + 1 = 0.$
Answer$2x^2 – 3x + 1 = 0; x = -1.$
Putting $x = -1$ in $L.H.S$. of equation
$L.H.S. = 2(-1)^2 - 3 x -1 + 1$
$= 2 + 3 + 1$
$= 6 \neq 0 \neq R.H.S.$
Hence, $x = -1$ is not a root of the equation.
View full question & answer→Question 202 Marks
Without actually determining the roots comment upon the nature of the roots of each of the following equations:
$x^2+ 5x + 15 = 0.$
Answer$x^2+ 5x + 15 = 0$
Here, $aa= 1, b = 5$ and $c = 15$
$D = b^2 - 4ac$
$= (5)^2- 4 x 1 x 15$
$= 25 - 60$
$= -35$
$\Rightarrow D < 0$
roots are imaginary.
View full question & answer→Question 212 Marks
Without actually determining the roots comment upon the nature of the roots of each of the following equations:
$x^2- 4x + 1 = 0$
Answer$x^2- 4x + 1 = 0$
Here, $a = 1, b = -4$ and $c = 1$
$D = b^2 - 4ac$
$\Rightarrow 16 - 4 x 1 x 1$
$\Rightarrow 16 - 4$
$= 12 > 0$
The given equation has real roots.
View full question & answer→Question 222 Marks
Without actually determining the roots comment upon the nature of the roots of each of the following equations:
$x^2 - 5x + 7 = 0$
Answer$x^2 - 5x + 7 = 0$
Here, $a = 1, b = -5$ and $c = 7$
$D = b^2- 4ac$
$\Rightarrow 25 - 4 x 1 x 7.$
$\Rightarrow 25 - 28$
$= -3$
Since $D < 0$ roots are imaginary.
View full question & answer→Question 232 Marks
Without actually determining the roots comment upon the nature of the roots of each of the following equations:
$9a^2b^2x^2 - 48abc + 64c^2d^2 = 0, a \neq 0, b \neq 0$
Answer$9a^2b^2x^2 - 48abc + 64c^2d^2 = 0$
Here, $D = b^2 - 4ac$
$\Rightarrow (-48abcd)^2 - 4 x 9a^2b^2 x 64c^2d^2$
$2304a^2b^2c^2d^2 - 2304a^2b^2c^2d^2 = 0$
$D = 0$
Roots are real and equal.
View full question & answer→Question 242 Marks
Without actually determining the roots comment upon the nature of the roots of each of the following equations:
$2 \sqrt{3} x^2-2 \sqrt{2} x-\sqrt{3}=0$
Answer$2 \sqrt{3} x^2-2 \sqrt{2} x-\sqrt{3}=0$
Here, $a=2 \sqrt{3}, b=-2 \sqrt{2}$ and $c=-\sqrt{3}$
$D=b^2-4 a c $
$\Rightarrow D-8-4 \times 2 \sqrt{3} \times-\sqrt{3} $
$\Rightarrow D=8+24 \\ =32>0$
The given equation has real roots.
View full question & answer→Question 252 Marks
Without actually determining the roots comment upon the nature of the roots of each of the following equations:
$3x^2 + 2x - 1 = 0$
Answer$3x^2 + 2x - 1 = 0$
Here, $a = 3, b = 2$ and $c = -1$
$D = b^2 - 4ac$
$= 4 - 4 x 3 x (-1)$
$\Rightarrow D = 4 + 12$
$= 16 > 0.$
View full question & answer→Question 262 Marks
Find the value of $k$ for which the given equation has real roots:
$9x^2 + 3kx + 4 = 0$.
AnswerThe given quadratic equation is :
$9 x^2+3 k x+4=0$
Here, a = 9, b = 3k and c = 4
This equation has real roots if
$b^2-4 a c \geq 0$
$ \Rightarrow(3 k)^2-4 \times 9 \times 4 \geq 0$
$ \Rightarrow 9 k^2-144 \geq 0 $
$ \Rightarrow 9 k^2 \geq 144 $
$ \Rightarrow k^2 \geq \frac{144}{9} $
$ \Rightarrow k \geq \frac{12}{3}$.
$\Rightarrow k \geq 4 $
View full question & answer→Question 272 Marks
Find the value of $k$ for which the given equation has real roots:
$kx^2 - 6x - 2 = 0$
AnswerThe given equation is :
$k x^2-6 x-2=0$
Here, $a = k, b = -6 \ \&\ c = -2$
This equation has real root if
$b^2-4 a c \geq 0 $
$ \Rightarrow(-6)^2-4 \times k \times(-2) \geq 0$
$ \Rightarrow 36+8 k \geq 0 $
$ \Rightarrow 8 k \geq-36 $
$ \Rightarrow k \geq-\frac{36}{8} $
$ \Rightarrow k \geq-\frac{9}{2} $
View full question & answer→Question 282 Marks
Form the quadratic equation whose roots are:
$\sqrt{3}$ and $3 \sqrt{3}$
AnswerLet a, β be the roots of the required quadratic equation:
Then, $a =\sqrt{3}$ and $\beta=3 \sqrt{3}$
$a+\beta=\sqrt{3}+3 \sqrt{3} \text { and } a \beta=\sqrt{3} \times 3 \sqrt{3} $
$\therefore a+\beta=4 \sqrt{3} \text { and } a \beta=9$
Required quadratic equation
$x^2-(a+\beta) x+a \beta=0$
$ \Rightarrow x^2-4 \sqrt{3} x+9=0$
View full question & answer→Question 292 Marks
$\left(x-\frac{a}{b}\right)^2=\frac{a^2}{b^2}$
Answer$\left(x-\frac{a}{b}\right)^2=\frac{a^2}{b^2} $
$ \Rightarrow x^2+\frac{a^2}{b^2}-\frac{2 a x}{b}=\frac{a^2}{b^2} $
$ \Rightarrow x^2-\frac{2 a x}{b}=0 $
$ \Rightarrow x\left(x-\frac{2 a}{b}\right)=0 $
$ \Rightarrow x =0 \text { or } x -\frac{2 a}{b}=0 $
$ \Rightarrow x =0 \text { or } x =\frac{2 a}{b} $
View full question & answer→Question 302 Marks
$3a^2x^2 + 8abx + 4b^2 = 0$
Answer$3 a^2 x^2+8 a b x+4 b^2=0 $
$ \Rightarrow 3 a^2 x^2+6 a b x+2 a b x+4 b^2=0$
$\Rightarrow 3 a x(a x+2 b)+2 b(a x+2 b)=0$
$ \Rightarrow(3 a x+2 b)(a x+2 b)=0$
$ \Rightarrow x=-\frac{2 b}{3 a} \text { or }-\frac{2 b}{a} $
View full question & answer→Question 312 Marks
Examine whether the equation $5x^2 -6x + 7 = 2x^2– 4x + 5$ can be put in the form of a quadratic equation.
Answer$5x^2 - 6x + 7 = 2x^2– 4x + 5$
$\Rightarrow 5x^2- 6x + 7 - 2x^2 + 4x - 5 = 0$
$\Rightarrow 3x^2 - 2x + 2 = 0.$
View full question & answer→Question 322 Marks
In each of the following determine the; value of k for which the given value is a solution of the equation:
$x^2 + 2ax - k = 0; x = - a.$
AnswerSince, $x = -a$ is a root of the equation
$x^2 + 2ax - k = 0$
$\Rightarrow (-a)^2 + 2a x (-a) - k = 0$
$\Rightarrow a^2 - 2a^2 - k = 0$
$\Rightarrow -k = a^2$
$\Rightarrow k = -a^2.$
View full question & answer→Question 332 Marks
In each of the following determine the; value of k for which the given value is a solution of the equation:
$3 x^2+2 k x-3=0 ; x=-\frac{1}{2}$
AnswerSince, x = $-\frac{1}{2}$ is a root of the given equation $3x^2+ 2kx - 3 = 0$
Therefore,
$3\left(-\frac{1}{2}\right)^2+2 k\left(-\frac{1}{2}\right)-3=0 $
$ \Rightarrow 3 \times \frac{1}{4}-k-3=0 $
$ \Rightarrow k =\frac{3}{4}-3 $
$=-\frac{9}{4} $
$\Rightarrow k =-\frac{9}{4}$
View full question & answer→Question 342 Marks
In each of the following determine the; value of k for which the given value is a solution of the equation:
$kx^2 + 2x - 3 = 0; x = 2$
AnswerSince, $x = 2$ is a root of the given equation, therefore, it satisfies the equation i.e.,
$ k (2)^2+2 \times 2-3=0 $
$ \Rightarrow 4 k +1=0 $
$ \Rightarrow k =-\frac{1}{4} $
View full question & answer→Question 352 Marks
The length of verandah is $3m$ more than its breadth. The numerical value of its area is equal to the numerical value of its perimeter.
(i) Taking x, breadth of the verandah write an equation in ‘x’ that represents the above statement.
(ii) Solve the equation obtained in above and hence find the dimension of verandah.
AnswerLet breadth = xm, length $= (x + 3)m.$
Area $= x (x + 3)$ sq.m.
Perimeter$ = 2(x + x + 3) = (4x + 6)m.$
According to the question, $x(x + 3) = 4x + 6$
$\Rightarrow x^2 - x - 6 = 0$
$\Rightarrow (x + 2) (x ++ 3) = 0$
$\therefore x = 3$ and $x = -2$ (inadmissiable).
Hence breadth = 3m, length = 6m.
View full question & answer→Question 362 Marks
Which of the following are quadratic equation :
$\left(x-\frac{1}{x}\right)^2=0$.
AnswerGiven equation is
$\left(x-\frac{1}{x}\right)^2=0$
$\Rightarrow x^2+\frac{1}{x^2}-2 x \frac{1}{x}=0 $
$\Rightarrow x^4+1-2 x^2=0 $
$ \Rightarrow x^4-2 x^2+1=0$
It is not a quadratic equation.
View full question & answer→Question 372 Marks
Which of the following are quadratic equation :
$(2x - 3) (x + 5) = 2 - 3x$
AnswerGiven equation
$(2x - 3) (x + 5) = 2 - 3x$
$\Rightarrow 2x^2 + 10 x - 3x - 15 - 2 + 3x = 0$
$\Rightarrow 2x^2 + 10x - 17 = 10$
It is quadratic equation.
View full question & answer→Question 382 Marks
$\sqrt{3} x^2+10 x+7 \sqrt{3}=0$
Answer$\sqrt{3} x^2+10 x+7 \sqrt{3}=0$
$\Rightarrow \sqrt{3} x^2+3 x+7 x+7 \sqrt{3}=0 $
$ \Rightarrow \sqrt{3} x(x+\sqrt{3})+7(x+\sqrt{3})=0$
$\Rightarrow \sqrt{3} x+7=0 \text { or } x +\sqrt{3}=0$
$x=-\frac{7}{\sqrt{3}}$ or $-\sqrt{3}$ are two roots of equation.
View full question & answer→