Question 12 Marks
Which of the following are quadratic equations?
$(x+2)^3=x^3-4$
AnswerHere it has been given that,
$(x+2)^3=x^3-4$
Now, after solving the above equation further we get
$ x^3+8+3(x)(2)(x+2)=x^3-4 $
$12+6 x^2+12 x=0 $
$ x^2+2 x+2=0$
Now as we can see, the above equation clearly represents a equation os the form $ax^2+ bx + c = 0$, where $a = 1, b= 2$ and $c = 2.$
Hence the above equation is a quadratic equation.
View full question & answer→Question 22 Marks
In the following, find the value of k for which the given value is a solution of the given equation:
$ x^2+3 a x+k=0, x=-a $
Answer$ x^2+3 a x+k=0, x=-a $
$ \because x=-a \text { is the solution } $
$ \therefore(-a)^2+3 a(-a)+k=0 $
$ \Rightarrow a^2-3 a^2+k=0 $
$ \Rightarrow-2 a^2+k=0 $
$ \therefore k=2 a^2$
View full question & answer→Question 32 Marks
Write the discriminant of the following quadratic equation.
$2 x^2-5 x+3=0$
Answer$2 x^2-5 x+3=0$
$\text { Here } a=2, b=-5, c=3$
$ \therefore \text { Discriminate }=b^2-4 a c $
$ =(-5)^2-4 \times 2 \times 3 $
$ =25-24 $
$ =1$
View full question & answer→Question 42 Marks
Which of the following are quadratic equations?
$\text{x}+\frac{1}{\text{x}}=1$
AnswerHere it has been given that,
$\text{x}+\frac{1}{\text{x}}=1$
Now, solving the above equation further we get,
$\frac{\text{x}^2+1}{\text{x}}=1$
$x^2+ 1 = 0$
$x^2- x + 1 = 0$
Now as can seen, the above equation clearly represents a quadratic equation of the form $ax^2+ bx + c = 0 $ where $a = 1, b = -1$ and $c = 1.$
Hence, the above equation is a quadratic equation.
View full question & answer→Question 52 Marks
Write the discriminant of the following quadratic equation.
$(x - 1)(2x - 1) = 0$
Answer$(x - 1)(2x - 1) = 0$
$2x^2- x - 2x + 1= 0$
$⇒ 2x^2- 3x + 1 = 0$
Here $a = 2, b = -3, c = 1$
$\therefore$ Discriminate $= b^2- 4ac$
$= (-3)^2- 4 × 2 × 1$
$= 9 - 8 = 1$
View full question & answer→Question 62 Marks
Solve the following quadratic equations by factorization:
$(x - 4)(x + 2) = 0$
AnswerWe have
$(x - 4)(x + 2) = 0$
$⇒$ either $(x - 4) = 0$ or $(x + 2) = 0$
$⇒ x = 4$ or $x = -2$
Thus, $x = 4$ and $x = -2$ are two roots of the equation $(x - 4)(x + 2) = 0$
View full question & answer→Question 72 Marks
Determine the nature of the root of the following quadratic equation:
$\frac{3}{5}\text{x}^2-\frac{2}{3}\text{x}+1=0$
Answer$\frac{3}{5}\text{x}^2-\frac{2}{3}\text{x}+1=0$
The given quadratic equation is $\frac{3}{5}\text{x}^2-\frac{2}{3}\text{x}+1=0$ can also be written as $9x^2- 10x + 15 = 0$
Here, $a = 9, b = -10, c = 15$
$\Rightarrow D=b^2-4 a c $
$ \Rightarrow D=(-10)^2-4 \times 15 \times 9$
$⇒ D = 100 - 540$
$⇒ D = -440 < 0$
$\therefore$ as $D > 0,$ the equation has no real roots.
View full question & answer→Question 82 Marks
Determine the nature of the root of the following quadratic equation:
$3\text{x}^2-2\sqrt{6}\text{x}+2=0$
Answer$3\text{x}^2-2\sqrt{6}\text{x}+2=0$
Here $\text{a}=3,\text{b}=-2\sqrt{6},\text{c}=2$
$\therefore$ Discriminant $(\text{D})=\text{b}^2-4\text{ac}$
$=(-2\sqrt{6})^2-4\times3\times2$
$=24-24=0$
$\because\text{D}=0$
$\therefore$ Roots are real and equal.
View full question & answer→Question 92 Marks
In the following, determine whether the given quadratic equation have real root and if so, find the root:
$x^2+x+2=0$
Answer$x^2+x+2=0$
$\text { The given equation is in the form of } a x^2+b x+c=0$
$ a=1, b=1, c=2 $
$ D=b^2-4 a c $
$ =(1)^2-4 \times 1 \times 2$
$= 1 - 8$
$= -7 < 0$
As $Q < 0,$ the equation has no real root.
View full question & answer→Question 102 Marks
Write the discriminant of the following quadratic equation.
$\sqrt{3}\text{x}^2+2\sqrt{2}\text{x}-2\sqrt{3}=0$
Answer $\sqrt{3}\text{x}^2+2\sqrt{2}\text{x}-2\sqrt{3}=0$
Here $\text{a}=\sqrt{3},\text{b}=2\sqrt{2},\text{c}=-2\sqrt{3}$
$\therefore$ Discriminate $=\text{b}^2-4\text{ac}$
$=(2\sqrt{2})^2-4\times\sqrt{3}\times(-2\sqrt{3})$
$=8+24=32$
View full question & answer→Question 112 Marks
Determine the nature of the root of the following quadratic equation:
$3\text{x}^2-4\sqrt{3}\text{x}+4=0$
AnswerThe given quadric is $3\text{x}^2-4\sqrt{3}\text{x}+4=0$
Here, $\text{a}=3,\text{b}=-4\sqrt{3}$ and $\text{c}=4$
As we know that $\text{D}=\text{b}^2-4\text{ac}$
Putting the value of $\text{a}=3,\text{b}=-4\sqrt{3}$ and $\text{c}=4$
$=(-4\sqrt{3})^2-4\times3\times4$
$=48-48$
$=0$
Since, $D = 0$
Therefore, root of the given equation are real and equal.
View full question & answer→Question 122 Marks
Which of the following are quadratic equations$?$
$\text{x}+\frac{1}{\text{x}}=\text{x}^2,\text{x}\neq0$
AnswerHere is has been given taht,
$\text{x}+\frac{1}{\text{x}}=\text{x}^2$
Now, solving the above equation further we get,
$\Big(\frac{\text{x}^2+1}{\text{x}}\Big)=\text{x}^2$
$x^2+1=x^3$
$ -x^3+x^2+1=0$
Now as we can see, the above equation clearly does not respresent a quadratic equation of the from $ax^2+ bx + c = 0,$ because $-x^3+ x^2+ 1$ is a polynomial having a degree of $3$ which is never present in a quadratic polynomial.
Hence, the above equation is not quadratic equation.
View full question & answer→Question 132 Marks
Write the discriminant of the following quadratic equation.
$x^2+ 2x + 4 = 0$
AnswerThe given equation is in the form of $a x^2+b x+c=0$
Here $a = 1, b = 2$ and $c = 4$
The discriminant is $D=b^2-4 a c$
$\Rightarrow(2)^2-4 \times 1 \times 4$
$⇒ 4 - 16 = -12$
$\therefore$ The discriminant of the following quadratic equation is $-12$
View full question & answer→Question 142 Marks
What is the nature of root of the quadratic equation $ 4 x^2-12 x-9=0 $.
Answer$ 4 x^2-12 x-9=0 $
$ \text { Here } a=4, b=-12, c=-9 $
$ \text { Discriminant }(D)=b^2-4 a c $
$ =(-12)^2-4 \times 4 \times(-9) $
$ =144+144=288 $
$ =D>0$
Roots are real and distinct.
View full question & answer→Question 152 Marks
Which of the following are quadratic equations?
$x^2-3 x=0$
Answer$x^2-3 x=0$
$\because$ $x^2- 3x$ is a quadratic polynomial.
$\therefore$ It is a quadratic equation.
View full question & answer→Question 162 Marks
Which of the following are quadratic equations?
$\text{x}^2-2\text{x}-\sqrt{\text{x}}-5=0$
AnswerHere it has been given that,
$\text{x}^2-2\text{x}-\sqrt{\text{x}}-5=0$
Now, as we can see the equation clearly does not represent a quadratic of the form $ax^2+ bx + c = 0,$ because $\text{x}^2-2\text{x}-\sqrt{\text{x}}-5=0$ contains an extra term $\text{x}^{\frac{1}{2}}$where $\frac{1}{2}$ is not an integer.
Hence, the above equation is not quadratic equation.
View full question & answer→Question 172 Marks
Determine the nature of the root of the following quadratic equation:
$2 x^2-6 x+3=0$
Answer$2 x^2-6 x+3=0$
$\text { Here, } a=2, b=-6, c=3$
$ \therefore D=b^2-4 a c $
$ \Rightarrow D=(-6)^2-4 \times 2 \times 3 $
$ \Rightarrow D=36-24 $
$ \Rightarrow D=12 $
$ \because D>0$
$\therefore$ Roots are real and distinct.
View full question & answer→Question 182 Marks
Find the discriminant of the quadratic equation $3\sqrt{3}\text{x}^2+10\text{x}+\sqrt{3}=0.$
Answer $3\sqrt{3}\text{x}^2+10\text{x}+\sqrt{3}=0$
Here $\text{a}=3\sqrt{3},\text{b}=10,\text{c}=\sqrt{3}$
$\therefore$ Discriminant $\text{D}=\text{b}^2-4\text{ac}$
$\text{D}=(10)^2-4\times3\sqrt{3}\times\sqrt{3}$
$\text{D}=100-12\times3$
$\text{D}=100-36$
$\text{D}=64$
View full question & answer→Question 192 Marks
Which of the following are quadratic equations?
$\text{x}^2+\frac{1}{\text{x}^2}=5$
Answer$\text{x}^2+\frac{1}{\text{x}^2}=5$
$\Rightarrow\text{x}^4-5\text{x}^2+1=0$
$\because\text{x}^4-5\text{x}^2+1=0$ is a $4$ degree polynomial
$\therefore$ it is a not a quadratic equation.
View full question & answer→Question 202 Marks
In the following, find the value of k for 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 $
$ \because x=a \text { is its solution } $
$ \therefore(a)^2-a(a+b)+k=0 $
$ \Rightarrow a^2-a^2-a b+k=0 $
$ \Rightarrow-a b+k=0 $
$ \therefore k=a b$
View full question & answer→Question 212 Marks
Which of the following are quadratic equations?
$ 16 x^2-3=(2 x+5)(5 x-3) $
Answer$ 16 x^2-3=(2 x+5)(5 x-3) $
$ \Rightarrow 16 x^2-3=10 x^2-6 x+25 x-15 $
$ \Rightarrow 16 x^2-3-10 x^2+6 x-25 x+15=0 $
$ \Rightarrow 6 x^2-19 x+12=0$
$\because 6 x^2-19 x+12 \text { is a quadratic polynomial. }$
$\therefore$ it is a quadratic equation.
View full question & answer→Question 222 Marks
Which of the following are quadratic equations?
$x^2+6 x-4=0$
Answer$x^2+6 x-4=0$
$\because$ $x^2+6 x-4=0$ is a quadratic polynomial
$\therefore$ It is a quadratic equation.
View full question & answer→Question 232 Marks
A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be $55$ minus the number of articles produced in a day. On a particular day, the total cost of production was $Rs. 750$. If $x$ denotes the number of toys produced that day, from the quadratic equation of find $x.$
AnswerLet the number to toys in a day $= x$
Cost of each toy $= x - 55$
On a particular cost of production $= Rs.\ 750$
$x(x - 55) = 750$
$\Rightarrow x^2-55 x-750=0$
Hence required quadratic equation will be $=x^2-55 x-750=0$
View full question & answer→Question 242 Marks
Which of the following are quadratic equations?
$2\text{x}^2-\sqrt{3}\text{x}+9=0$
Answer$2\text{x}^2-\sqrt{3}\text{x}+9=0$
$\because2\text{x}^2-\sqrt{3}\text{x}+9=0$ is a quadratic polynomial.
$\therefore$ it is a quadratic equation.
View full question & answer→Question 252 Marks
Solve the following quadratic equations by factorization:
$3x^2- 14x - 5 = 0$
Answer$ 3 x^2-14 x-5=0 $
$ \Rightarrow 3 x^2-15 x+x-5=0$
$\begin{cases}\because -5\times3=-15& \\\therefore-15=-15\times1\\-14=-15+1\end{cases}$
$⇒ 3x(x - 5) + 1(x - 5) = 0$
$⇒ (x - 5)(3x + 1) = 0$
Either $x - 5 = 0$, then $x = 5$ or $3x + 1 = 0,$
Then $3x = -1$
$\Rightarrow\text{x}=\frac{-1}{3}$
Roots are $x = 5, \frac{-1}{3}$
View full question & answer→Question 262 Marks
Find the value of k for which the following equation have real root:
$2 x^2+k x+3=0$
Answer$2 x^2+k x+3=0$
Here $a=2, b=k, c=3$
$ \therefore D=b^2-4 a c $
$ D=k^2-4 \times 2 \times 3 $
$ D=k^2-24$
$\because$ Roots are real and equal
$ \therefore D=0 $
$ \Rightarrow k^2-24=0 $
$ \Rightarrow k^2=24$
$\text{k}=\pm\sqrt{24}$
$\text{k}=\pm\sqrt{4\times6}$
$\text{k}=\pm2\sqrt6$
View full question & answer→Question 272 Marks
Which of the following are quadratic equations$?$
$\text{x}-\frac{3}{\text{x}}=\text{x}^2$
AnswerHere it has been given that,
$\text{x}-\frac{3}{\text{x}}=\text{x}^2$
Now, solving the above equation further we get
$\frac{\text{x}^2-3}{\text{x}}=\text{x}^2$
$-x^3+x^2-3=0$
Now, the above equation clearly does not represent a quadratic equation of the form $a x^2+b x+c=0$, because $-x^3+x^2-3$ is a polynomial of degree $3 .$
Hence, the above equation is not a quadratic equation.
View full question & answer→Question 282 Marks
Which of the following are quadratic equations$?$
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=3\Big(\text{x}+\frac{1}{\text{x}}\Big)+4$
AnswerHere it has been given that,
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=3\Big(\text{x}+\frac{1}{\text{x}}\Big)+4$
Now, solving the above equation further we get,
$\Big(\frac{\text{x}^2+1}{\text{x}}\Big)^2=3\Big(\frac{\text{x}^2+1}{\text{x}}\Big)+4$
$\frac{(\text{x}^2+1)^2}{\text{x}^2}=\frac{3\text{x}^2+3+4\text{x}}{\text{x}}$
$\text{x}^4 + 1 + 2\text{x}^2 = 3\text{x}^3 + 4\text{x}^2 + 3\text{x}$
$\text{x}^4 - 3\text{x}^3 - 2\text{x}^2 - 3\text{x} + 1 = 0$
Now as we can see, the above equation clearly does not respresent a quadratic equation of the from $ax^2+ bx + c = 0,$ because $a^4- 3x^2- x + 1$ is a polynomial having a degree of $4$ which is never present in a quadratic polynomial.
Hence, the above equation is not a quadratic equation.
View full question & answer→Question 292 Marks
Write the discriminant of the following quadratic equation.
$x^2- 2x + k = 0$, $\text{k}\in\text{R}$
AnswerThe given equation is in the form of $ax^2+ bx + c = 0$
Here $a = 1, b = -2, c = k$ $[$given $\text{k}\in\text{R}]$
The discriminant is $D = b^2- 4ac$
$⇒ (-2)^2- 4 × 1 × k$
$⇒ 4 - 4k$
The discriminant D, of the following quadratic equation is $4 - 4k,$ where $\text{k}\in\text{R}$
View full question & answer→Question 302 Marks
In the following, determine whether the given quadratic equation have real root and if so, find the root:
$ 3 x^2-2 x+2=0 $
Answer$ 3 x^2-2 x+2=0 $
$ \text { Here } a=3, b=-2, c=2 $
$ \therefore \text { Discriminate }=b^2-4 a c $
$ =(-2)^2-4 \times 3 \times 2 $
$ =4-24=-20 $
$ \because D<0 $
$ \therefore \text { Roots are not real. }$
View full question & answer→Question 312 Marks
Which of the following are quadratic equations?
$ 3 x^2-5 x+9=x^2-7 x+3 $
Answer$ 3 x^2-5 x+9=x^2-7 x+3 $
$ \Rightarrow 3 x^2-5 x+9-x^2+7 x-3=0 $
$ \Rightarrow 2 x^2+2 x+6=0$
$\because 2 x^2+2 x+6 \text { is a quadratic polynomial. }$
$\therefore$ It is a quadratic equation.
View full question & answer→Question 322 Marks
Write the discriminant of the following quadratic equation.
$\sqrt{3}\text{x}^2+2\sqrt{2}\text{x}-2\sqrt{3}=0$
Answer$\sqrt{3}\text{x}^2+2\sqrt{2}\text{x}-2\sqrt{3}=0$
Here $\text{a}=\sqrt{3},\text{b}=2\sqrt{2},\text{c}=-2\sqrt{3}$
$\therefore$ Discriminate $=\text{b}^2-4\text{ac}$
$=(2\sqrt{2})^2-4\times\sqrt{3}\times(-2\sqrt{3})$
$=8+24=32$
View full question & answer→Question 332 Marks
Solve the following quadratic equations by factorization:
$(2x + 3)(3x - 7) = 0$
AnswerWe have been given
$(2x + 3)(3x - 7) = 0$
Therefore,
$(2x + 3) = 0$
$2x = -3$
$\text{x}=\frac{-3}{2}$
or $(3x - 7) = 0$
$3x = 7$
$\text{x}=\frac{7}{3}$
Therefore, $\text{x}=\frac{-3}{2}$ or $\text{x}=\frac{7}{3}$
View full question & answer→Question 342 Marks
Which of the following are quadratic equations$?$
$(2x + 1)(3x + 2) = 6(x - 1)(x - 2)$
Answer$(2x + 1)(3x + 2) = 6(x - 1)(x - 2)$
$ \Rightarrow 6 x^2+4 x+3 x+2=6\left(x^2-2 x-x+2\right)$
$ \Rightarrow 6 x^2+7 x+2=6 x^2-18 x+12$
$ \Rightarrow 6 x^2+7 x+2-6 x^2+18 x-12=0 $
$ \Rightarrow 25 x-10=0$
$\because 25x - 10$ is one degree polynomial.
$\therefore$ It is not a quadratic equation.
View full question & answer→Question 352 Marks
Show that $x = -3$ is a solution of $x^2+6 x+9=0$.
AnswerThe given equation is $x^2+6 x+9=0$
If $x=-3$ is its solution then it will satisfy it.
$ \text { L.H.S. }=(-3)^2+6(-3)+9 $
$ =9-18+9 $
$ =18-18 $
$ =0 $
$ =\text { R.H.S. }$
Hence $x=-3$ is its one root.
View full question & answer→Question 362 Marks
Which of the following are quadratic equations?
$x(x + 1) + 8 = (x + 2)(x - 2)$
Answer$ x(x+1)+8=(x+2)(x-2) $
$ x^2+x+8=x^2-4 $
$ \Rightarrow x^2+x+8-x^2+4=0 $
$ \Rightarrow x+12=0$
$\because x + 12$ is not a quadratic polynomial.
$\therefore$ It is not a quadratic equation.
View full question & answer→Question 372 Marks
Which of the following are quadratic equations?
$\sqrt{3}\text{x}^2-2\text{x}+\frac{1}{2}=0$
AnswerHere it has been given that,
$\sqrt{3}\text{x}^2-2\text{x}+\frac{1}{2}=0$
Now, solving the above equation further we get,
$\frac{2\sqrt{3}\text{x}^2-4\text{x}+1}{2}=0$
$2\sqrt{3}\text{x}^2-4\text{x}+1=0$
Now, the above equation clearly represents a quadratic equation of the form $ax^2+ bx + c = 0,$ where $\text{a}=2\sqrt{3},$ b = -4 and c = 1.
Hence, the above equation is a quadratic equation.
View full question & answer→Question 382 Marks
Write the discriminant of the following quadratic equation.
$x^2-x+1=0$
Answer$x^2-x+1=0$
The given equation is in form of $a x^2+b x+c=0$
Here, $a=1, b=-1$ and $c=1$
The discriminant is $D=b^2-4 a c$
$ \Rightarrow(-1)^2-4 \times 1 \times 1 $
$ \Rightarrow 1-4=-3$
$\therefore$ The discriminant $D$, of the following quadratic equation is $-3$
View full question & answer→Question 392 Marks
Determine the nature of the root of the following quadratic equation:
$2 x^2-3 x+5=0$
AnswerThe given quadric equation is $2 x^2-3 x+5=0$
Here, $\mathrm{a}=2, \mathrm{~b}=-3$ and $\mathrm{c}=5$
As we know that $D=b^2-4 a c$
Putting the value of $a=2, b=-3$ and $c=5$
$ =(-3)^2-4 \times 2 \times 5$
$=9-40$
$ =-31$
Since, $D < 0$
Therefore, root of the given equation are not real.
View full question & answer→Question 402 Marks
Write the set of values of $k$ for which the quadratic equation has $2x^2+ kx - 8 = 0$ has real roots.
Answer$ \text { In } 2 x^2+k x-8=0$
$ D=b^2-4 a c $
$ D=(k)^2-4 \times 2 \times(-8)$
$ D=k^2+64$
The roots are real
$\text{D}\geq0$
$\text{k}^2+64\geq0$
For all real value of $k,$ the equation has real roots.
View full question & answer→