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
$x^3 + 8 + 3(x)(2)(x + 2) = x^3 - 4$
$12 + 6x^2 + 12x = 0$
$x^2 + 2x + 2 = 0$
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:
$x^2 + 3ax + k = 0, x = -a$
Answer$x^2 + 3ax + k = 0, x = -a$
$\because$ $x = -a$ is the solution
$\therefore$ $(-a)^2+ 3a(-a) + k = 0$
$\Rightarrow a^2 - 3a^2 + k = 0$
$\Rightarrow -2a^2 + k = 0$
$\therefore$ $k = 2a^2$
View full question & answer→Question 32 Marks
Write the discriminant of the following quadratic equation.
$2x^2 - 5x + 3 = 0$
Answer$2x^2 - 5x + 3 = 0$
Here $a = 2, b = -5, c = 3$
$\therefore$ Discriminate $= b^2 - 4ac$
$= (-5)^2- 4 \times 2 \times 3$
$= 25 - 24$
$= 1$
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$(x - 1)(2x - 1) = 0$
$2x^2 - x - 2x + 1= 0$
$\Rightarrow 2x^2 - 3x + 1 = 0$
Here $a = 2, b = -3, c = 1$
$\therefore$ Discriminate $= b^2 - 4ac$
$= (-3)^2 - 4 \times 2 \times 1$
$= 9 - 8 = 1$
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, $a = 9, b = -10, c = 15$
$\Rightarrow D = b^2 - 4ac$
$\Rightarrow D = (-10)^2 - 4 \times 15 \times 9$
$\Rightarrow D = 100 - 540$
$\Rightarrow D = -440 < 0$
$\therefore$ 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$x^2 + x + 2 = 0$
The given equation is in the form of $ax^2 + bx + c = 0$
$a = 1, b = 1, c = 2$
$D = b^2 - 4ac$
$= (1)^2 - 4 \times 1 \times 2$
$= 1 - 8$
$= -7 < 0$
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 $ax^2 + bx + c = 0$
Here $a = 1, b = 2$ and $c = 4$
The discriminant is $D = b^2 - 4ac$
$\Rightarrow (2)^2 - 4 \times 1 \times 4$
$\Rightarrow 4 - 16 = -12$
$\therefore$ 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 $4x^2 - 12x - 9 = 0.$
Answer$4x^2 - 12x - 9 = 0$
Here $a = 4, b = -12, c = -9$
Discriminant $(D) = b^2 - 4ac$
$= (-12)^2 - 4 \times 4 \times (-9)$
$= 144 + 144 = 288$
$= D > 0$
Roots are real and distinct.
View full question & answer→Question 152 Marks
Which of the following are quadratic equations?
$x^2 - 3x = 0$
Answer$x^2 - 3x = 0$
$\because$ $x^2 - 3x$ 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:
$2x^2 - 6x + 3 = 0$
Answer$2x^2 - 6x + 3 = 0$
Here, $a = 2, b = -6, c = 3$
$\therefore$ $D = b^2 - 4ac$
$\Rightarrow D = (-6)^2 - 4 \times 2 \times 3$
$\Rightarrow D = 36 - 24$
$\Rightarrow D = 12$
$\because$ $D > 0$
$\therefore$ 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 is its solution
$\therefore$ $(a)^2 - a(a + b) + k = 0$
$\Rightarrow a^2 - a^2 - ab + k = 0$
$\Rightarrow -ab + k = 0$
$\therefore$ $k = ab$
View full question & answer→Question 212 Marks
Which of the following are quadratic equations?
$16x^2 - 3 = (2x + 5)(5x - 3)$
Answer$16x^2 - 3 = (2x + 5)(5x - 3)$
$\Rightarrow 16x^2 - 3 = 10x^2 - 6x + 25x - 15$
$\Rightarrow 16x^2 - 3 - 10x^2 + 6x - 25x + 15 = 0$
$\Rightarrow 6x^2 - 19x + 12 = 0$
$\because$ $6x^2 - 19x + 12$ is a quadratic polynomial.
$\therefore$ it is a quadratic equation.
View full question & answer→Question 222 Marks
Which of the following are quadratic equations?
$x^2 + 6x - 4 = 0$
Answer$x^2 + 6x - 4 = 0$
$\because$ $x^2 + 6x - 4 = 0$ 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$
$x(x - 55) = 750$
$\Rightarrow x^2 - 55x - 750 = 0$
Hence required quadratic equation will be $x^2 - 55x - 750 = 0$
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$3x^2 - 14x - 5 = 0$
$\Rightarrow 3x^2 - 15x + x - 5 = 0$
$\begin{cases}\because -5\times3=-15& \\\therefore-15=-15\times1\\-14=-15+1\end{cases}$
$\Rightarrow 3x(x - 5) + 1(x - 5) = 0$
$\Rightarrow (x - 5)(3x + 1) = 0$
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:
$2x^2 + kx + 3 = 0$
Answer$2x^2 + kx + 3 = 0$ Here $a = 2, b = k, c = 3$
$\therefore$ $D = b^2 - 4ac$
$D = k^2 - 4 \times 2 \times 3$
$D = k^2 - 24$
$\because$ 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 $ax^2 + bx + c = 0$, 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 $D = b^2 - 4ac$
$\Rightarrow (-2)^2 - 4 \times 1 \times k$
$\Rightarrow 4 - 4k$
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:
$3x^2 - 2x + 2 = 0$
Answer$3x^2 - 2x + 2 = 0$
Here $a = 3, b = -2, c = 2$
$\therefore$ Discriminate $= b^2 - 4ac$
$= (-2)^2 - 4 \times 3 \times 2$
$= 4 - 24 = -20$
$\because$ $D < 0$
$\therefore$ Roots are not real.
View full question & answer→Question 312 Marks
Which of the following are quadratic equations?
$3x^2 - 5x + 9 = x^2 - 7x + 3$
Answer$3x^2 - 5x + 9 = x^2 - 7x + 3$
$\Rightarrow 3x^2 - 5x + 9 - x^2 + 7x - 3 = 0$
$\Rightarrow 2x^2+ 2x + 6 = 0$
$\because$ $2x^2+ 2x + 6$ 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$(2x + 1)(3x + 2) = 6(x - 1)(x - 2)$
$\Rightarrow 6x^2 + 4x + 3x + 2 = 6(x^2 - 2x - x + 2)$
$\Rightarrow 6x^2 + 7x + 2 = 6x^2 - 18x + 12$
$\Rightarrow 6x^2 + 7x + 2 - 6x^2 + 18x - 12 = 0$
$\Rightarrow 25x - 10 = 0$
$\because$ $25x - 10$ 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 $x^2 + 6x + 9 = 0.$
AnswerThe given equation is $x^2 + 6x + 9 = 0$
If $x = -3$ is its solution then it will satisfy it.
L.H.S. $= (-3)^2 + 6(-3) + 9$
$= 9 - 18 + 9$
$= 18 - 18$
$= 0$
= 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 $ax^2 + bx + c = 0$
Here, $a = 1, b = -1$ and $c = 1$
The discriminant is $D = b^2 - 4ac$
$\Rightarrow (-1)^2 - 4 \times 1 \times 1$
$\Rightarrow 1 - 4 = -3$
$\therefore$ 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:
$2x^2 - 3x + 5 = 0$
AnswerThe given quadric equation is $2x^2 - 3x + 5 = 0$
Here, $a = 2, b = -3$ and $c = 5$
As we know that $D = b^2 - 4ac$
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.
AnswerIn $2x^2 + kx - 8 = 0$
$D = b^2 - 4ac$
$D = (k)^2 - 4 \times 2 \times (-8)$
$D = k^2 + 64$
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→