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CBSE Questions for Class 10 Maths Quadratic Equations Quiz 1 - MCQExams.com
CBSE
Class 10 Maths
Quadratic Equations
Quiz 1
Solve the following equations.
$$x^4\,-\,3x^2\,+\,2\,=\,0$$, roots are
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$$x\,=\,\pm\,\sqrt3,\,\pm\,1$$
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$$x\,=\,\pm\,\sqrt2,\,\pm\,1$$
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$$x\,=\,\pm\,\sqrt2,\,\pm\,3$$
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none of the above
Explanation
Given equation is $$x^4 - 3x^2 +2 = 0$$
$$x^4 - 2x^2 -x^2 + 2 =0$$
$$(x^2 - 1)(x^2 - 2) = 0$$
$$\therefore x^2 = 2 \Rightarrow x = \pm \sqrt{2}$$
$$\therefore x^2 = 1 \Rightarrow x = \pm 1$$
Which of the following methods is used to derive the Standard Quadratic Formula for the Quadratic Equation $$ax^2+bx+c=0$$?
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Factorisation Method
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Completing Square Method
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Hit and Trial Method
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All the above.
Explanation
We have learnt that in order to derive the standard quadratic formula from the standard form of a quadratic equation $$ax^2+bx+c=0$$,we follow the steps involved in the Completing Square Method . Hence $$b$$ is the correct option.
Let $$\alpha(a)$$ and $$\beta(a)$$ be the roots of the equation $$\\ \left( \sqrt [ 3 ]{ 1+a } -1 \right) x^{ 2 }+\left( \sqrt { 1+a } -1 \right) x+\left( \sqrt [ 6 ]{ 1+a } -1 \right) =0$$
$$\displaystyle \lim _{ a\rightarrow 0 }{ \alpha } \left( a \right)$$ and $$\displaystyle \lim _{ a\rightarrow 0 }{ \beta } \left( a \right)$$ are
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$$-\dfrac{5}{2}$$ and $$1$$
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$$-\dfrac{11}{2}$$ and $$-1$$
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$$-\dfrac{7}{2}$$ and $$2$$
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$$-\dfrac{9}{2}$$ and $$3$$
The roots of the equation $$\sqrt{3y + 1} = \sqrt{y - 1}$$ are?
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$$- 1$$
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$$2, 3$$
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$$2, 1$$
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None of these
Explanation
Given: equation $$\sqrt {3y+1}=\sqrt {y-1}$$
To find the roots of the equation
Sol:
$$\sqrt {3y+1}=\sqrt {y-1}$$
Take square on both sides, we get
$$ {3y+1}= {y-1}\\\implies 3y-y=-1-1\\\implies 2y=-2$$
or, $$y=-1$$
But $$y\ne-1$$ as $$\sqrt{y-1} = \sqrt{-2}$$ which is not possible. Hence, none of the given options is the answer.
A quadratic equation in $$x$$ is $$ax^2 + bx + c = 0$$, where $$a, b, c$$ are real numbers and the other condition is
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$$a \neq 0$$
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$$b \neq 0$$
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$$c \neq 0$$
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$$b = 0$$
Explanation
$$a, b$$ and $$c$$ are constants in the equation.
As $$x$$ is raised to power $$2$$ it is a quadratic equation.
If $$a=0$$, then it will nullify $${ x }^{ 2 }$$, making it a linear equation.
So only if $$a\neq 0$$, the equation remains quadratic.
Option A is correct.
Find the roots of following quadratic equation
$$x^2\,+\,3x\,-\,2\,=\,0$$
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$$\displaystyle\,x\,=\,\frac{-3\,\pm\,\sqrt{15}}{2}$$
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$$\displaystyle\,x\,=\,\frac{-3\,\pm\,\sqrt{17}}{2}$$
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$$\displaystyle\,x\,=\,\frac{-3\,\pm\,\sqrt{19}}{2}$$
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$$\displaystyle\,x\,=\,\frac{-3\,\pm\,\sqrt{21}}{2}$$
Explanation
Gven equation is
$$x^2 + 3x - 2 = 0 $$
Using quadratic formula,
$$a=1, b=3, c=-2$$
$$x = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}$$
$$x = \frac{-3 \pm \sqrt{9 - 4(1)(-2)}}{2\times1}$$
$$\therefore x = \frac{-3 \pm \sqrt{17}}{2}$$
$$\therefore$$ the roots of the given equation are $$\displaystyle \frac {-3\pm \sqrt{17}}{2}$$
Find the discriminant for the given quadratic equation:
$$\sqrt{3}x^2\,+\,2\sqrt{2}x\,-\,2\sqrt{3}\,=\,0$$
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$$26$$
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$$32$$
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$$38$$
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$$44$$
Explanation
Given equation is,
$$\sqrt{3}x^2\,+\,2\sqrt{2}x\,-\,2\sqrt{3}\,=\,0$$
We know, $$D = b^2 - 4ac$$
$$a=\sqrt 3, b=2 \sqrt 2, c=-2\sqrt3$$
$$D = (2\sqrt{2})^2 - 4\times\sqrt{3}\times(-2\sqrt{3})$$
$$D = 8 + 24$$
$$\therefore D = 32$$
Find the discriminant for the given equation:
$$3x^2\,+\,2x\,-\,1\,=\,0$$
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$$11$$
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$$13$$
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$$15$$
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$$16$$
Explanation
Given equation is,
$$3x^2 + 2x- 1 = 0$$
We know, $$D = b^2 - 4c$$
$$a=3,b=2,c=-1$$
$$D = 2^2 - 4\times3\times(-1)$$
$$D = 4 + 12$$
$$\therefore D = 16$$
Find the discriminant for the given quadratic equation: $$x^2\,+\,4x\,+\,k\,=\,0$$
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$$6\,-\,4k$$
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$$16\,-\,4k$$
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$$16\,-\,3k$$
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$$13\,-\,2k$$
Explanation
Given equation is,
$$x^2 + 4x + k = 0 $$
We know,
$$D = b^2 - 4ac $$
$$a=1, b=4, c=k$$
$$D = (4)^2 - 4(1)(k)$$
$$\therefore D = 16 - 4k$$
Find the value of discriminant for the following equation.
$$x^{2}\, +\, 4x\, +\, k\, =\, 0$$
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$$\Delta\, =\, 16\, + 4k$$
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$$\Delta\, =\, 16\, -4k$$
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$$\Delta\, =\, 16\, - k$$
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$$\Delta\, =\, 16+k$$
Explanation
Given equation is:
$$x^2+4x+k=0$$
Discriminant = $$b^2 - 4ac$$
$$a=1, b=4, c=k$$
$$\therefore D= (4)^2 - 4(k)(1)$$
$$= 16 - 4k$$
$$\therefore \Delta =16-4k$$
Say true or false.
A quadratic equation cannot be solved by the method of completing the square.
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True
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False
Explanation
Given a quadratic equation is $$ax^2+bx+c=0$$
There are $$3$$ methods to solve the quadratic equations :
$$(1)$$ Factorisation(Find factors of the equation)
$$(2)$$ Using quadratic formula i.e $$x=\dfrac{-b\pm \sqrt{b^2 - 4ac}}{2a}$$
$$(3)$$ Completing the square
Thus, completing the square is one of the methods to solve quadratic equations.
Hence, the given statement is false.
The discriminant of $$ax^{2}-(a+b)x+b=0$$ is:
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$$a-b$$
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$$a+b$$
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$$(a+b)^{2}$$
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$$(a-b)^{2}$$
Explanation
(D) $$b^{2}-4ac=\left [ (a+b) \right ]^{2}-4ab$$
$$=(a+b)^{2}-4ab$$
$$= a^2+b^2-2ab$$
$$=(a-b)^{2}$$
If a given expression is a complete square, then which of the following formulae we use to factorize it?
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$$a^2 + 2ab+b^2=(a+b)^2$$
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$$a^2 - 2ab+b^2=(a-b)^2$$
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$$(a - b) (a + b) = (a^2 -b^2)$$
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$$(x+a)(x+b)=x^2+(a+b)x+ab$$
Explanation
If the given expression is a complete square, then we use the following formulae to factorize it.
$$a^2+2ab+b^2=(a+b)^2$$ and $$a^2-2ab+b^2=(a-b)^2$$
For the expression $$ax^2 + 7x + 2$$ to be quadratic, the necessary condition is
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$$a=0$$
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$$a \neq 0$$
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$$ a> \cfrac{7}{2}$$
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$$ a<-1 $$
Explanation
For the expression $$a^2+7x+2$$ to be quadratic the possible values of $$a$$ must be non zero real numbers because we know that the expression $$ax^2+bx+c$$ where $$a, b, c$$ are real numbers is quadratic if $$a\neq 0$$
Find the value of discriminant for the following equation.
$$2x^{2}\, +\, x\, +\, 1\, =\, 0$$
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$$7$$
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$$-7$$
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$$9$$
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$$-9$$
Explanation
The value of discriminant of $$2x^2 + x + 1 =0 $$ is
$$D = b^2 - 4ac$$
$$a=2, b=1, c=1$$
$$\therefore D = (1)^2 - 4\times2\times1 $$
$$= 1 - 8$$
$$= -7$$
The discriminant (D) of $$\sqrt{x^{2}+x+1}=2$$ is:
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-3
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13
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11
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12
Explanation
Squaring on the both sides of the given equation, we get
$$x^{2}+x+1=4$$
$$x^{2}+x-3=0$$
$$D=b^{2}-4ac$$
$$=1-4(1)(-3)$$
$$=1+12$$
$$=13$$
If the discriminant of $$3x^{2}-14x+k=0$$ is $$100$$, then $$k=$$
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$$8$$
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$$32$$
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$$16$$
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$$24$$
Explanation
Given that, the discriminant of $$3x^2-14x+k=0$$ is $$100$$.
To find out: The value of $$k$$.
We know that, the discriminant of a quadratic equation of the form $$ax^2+bx+c=0$$ is given by,
$$D=b^{2}-4ac$$
Here, $$a=3, b=-14, c=k$$
So, $$D=(-14)^{2}-4(3)k$$
$$\Rightarrow (-14)^{2}-4(3)k=100$$
$$\Rightarrow 196-12k=100$$
$$\Rightarrow 12k=96$$
$$\Rightarrow k=8$$
Hence, the value of $$k$$ is $$8$$.
Which of the following is a quadratic polynomial in one variable?
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$$\sqrt{2x^{3}}+5$$
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$$2x^2 + 2x^{-2}$$
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$$x^2$$
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$$2x^2+y^2$$
Explanation
Polynomials in one variable are algebraic expressions that consist of terms in the form $$ax^n$$
Here $$n$$ is a non-negative (i.e. positive or zero) integer and $$a$$ is a real number and is called the coefficient of the term.
The degree of a polynomial in one variable is the largest exponent in the polynomial.
And for quadratic polynomial $$n=2.$$
Hence the only quadratic polynomial in one variable is $$x^2$$
The mentioned equation is in which form?
$$3y^{2}\,-\, 7\, =\, \sqrt{3}\,y$$
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linear
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Quadratic
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Cubic
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None
Explanation
The highest power of y is 2. Hence, the equation is quadratic.
Which of the following is a quadratic equation ?
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$$x^{\frac{1}{2}}+2x+3=0$$
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$$(x^2-1)(x+4)=x^{2}+1$$
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$$x^{2}-3x+5=0$$
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$$(2x^2+1)(3x-4)=6x^{2}+3$$
Explanation
Standard form of a quadratic equation is $$ax^{2}+bx+c=0$$
The quadratic equation is an equation having $$2$$ highest degree of the variable, which is possible only in the equation $$x^2-3x+5=0$$ out of the four given equations.
Hence, $$Op-C$$ is correct.
The condition for $$px^{2}+qx+r=0$$ to be pure quadratic is:
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$$p=0$$
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$$q=0$$
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$$r=0$$
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$$p=q=0$$
Explanation
A
quadratic
equation in which the term containing x raised to the power of 1 is not present is called a
pure quadratic
equation. In other words, ax^2 + c = 0 is a
pure quadratic
equation.
$$ \therefore\ q\ = 0 $$
The mentioned equation is in which form?
$$\cfrac {3}{4}y^{2}\, =\, 2y\, +\,7$$
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cubic
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quadratic
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linear
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none of these
Explanation
Given equation is $$\dfrac {3}{4}y^2=2y+7$$
The highest power of $$x$$ is $$2$$. Thus, it is a quadratic equation.
The mentioned equation is in which form?
$$(y\, -\, 2)\, (y\, +\, 2)\, =\, 0$$
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cubic
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quadratic
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linear
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none of these
Explanation
Given, $$(y -2)(y +2) =0 $$
$$\Rightarrow y^2 - 4 = 0$$
The highest power of $$y$$ is $$2$$.
Thus, it is a quadratic equation.
The mentioned equation is in which form?
$$m^{3}\, +\, m\, +\, 2\, =\, 4m$$
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Linear
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Quadratic
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Quartic
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None
Explanation
No. The highest power of m is 3. Thus, it is not a quadratic equation.
The mentioned equation is in which form?
$$y^{2}\, -\, 4\, =\, 11y$$
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Quadratic
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linear
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Cubic
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None
Explanation
Given equation is $$y^2-4=11y$$
The highest power of $$y$$ is $$2$$. Thus, it is a quadratic equation.
The mentioned equation is in which form?
$$z\, -\, \cfrac{7}{z}\, =\, 4z\, +\, 5$$
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Linear
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Quadratic
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cubic
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None of these
Explanation
Given,
$$z - \dfrac{7}{z} = 4z + 5$$
$$\Rightarrow \dfrac { { z }^{ 2 }-7 }{ z } =4z+5$$
$$\Rightarrow z^2 - 7 = 4z^2 + 5z $$
$$\Rightarrow 3z^2 + 5z + 7 = 0 $$
The highest power of z is 2. Hence, it is a quadratic equation.
The mentioned equation is in which form?
$$n\, -\, 3\, =\, 4n$$
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linear
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Quadratic
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constant
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None
Explanation
No. The highest power of y is 1. Thus, it is not a quadratic equation.
STATEMENT - 1 : $$(x-2)(x+1)$$ $$=$$ $$(x-1)(x+3)$$ is a quadratic equation.
STATEMENT - 2 : If $$p(x)$$ is a quadratic polynomial, then $$p(x)$$ $$=$$ $$0$$ is called a quadratic equation.
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Statement - 1 is True, Statement - 2 is True, Statement - 2 is a correct explanation for Statement - 1
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Statement - 1 is True, Statement - 2 is True : Statement 2 is NOT a correct explanation for Statement - 1
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Statement - 1 is True, Statement - 2 is False
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Statement - 1 is False, Statement - 2 is True
Explanation
Given equation $$(x-2)(x+1)=(x-1)(x+3)$$
To find whether the given equation is a quadratic equation or not
Sol:
$$(x-2)(x+1)=(x-1)(x+3)\\\implies x^2+x-2x-2=x^2+3x-x-3\\\implies x^2-x^2+x-2x-3x+x-2+3=0\\\implies -3x+1=0$$
This is not a quadratic equation as an equation of degree two is called a quadratic equation.
And, a polynomial when equated to zero or some value becomes an equation.
Is the following equation quadratic?
$$(x\, +\, 3) (x\, -\, 4)\, =\, 0$$
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Yes
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No
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Ambiguous
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Data insufficient
Explanation
$$Answer=1$$
The equation $$(x\, +\, 3) (x\, -\, 4)\, =\, 0$$ can be formed as
$$x^2 - x -12 =0 $$ has the highest power as 2. Thus, it is quadratic equation.
Is the following equation quadratic?
$$n^{3}\, -\, n\, +\, 4\, =\, n^{3}$$
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Yes
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No
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Ambiguous
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Data insufficient
Explanation
A quadratic equation is a second-order polynomial equation in a single variable $$x$$.
$$ax^2+bx +c=0$$ where $$a\neq 0$$.
The equation $$n^3 - n + 4 = n^3$$ can be framed as $$-n + 4 =0 $$ by subtracting $${ n }^{ 3 }$$ on both sides,
Thus, it is not a quadratic equation as it is not an equation of degree $$2.$$
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