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CBSE Questions for Class 10 Maths Quadratic Equations Quiz 7 - MCQExams.com
CBSE
Class 10 Maths
Quadratic Equations
Quiz 7
Value(s) of $$x$$ which satisfies the equation $$x^{2} + 2kx = \dfrac {j}{3}$$, where $$j$$ and $$k$$ are constants, is/are
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$$x = -k\pm \dfrac {\sqrt {3(3k^{2} + j)}}{3}$$
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$$x = -6k\pm \dfrac {\sqrt {3(3k^{2} + j)}}{3}$$
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$$x = -k\pm \dfrac {\sqrt {3(3k^{2} + j)}}{6}$$
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$$x = -6k\pm \left (k + \dfrac {\sqrt {3j}}{3}\right )$$
Explanation
We know the quadratic equation formula,
$$\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$$
$$x^2+2kx=\dfrac{j}{3}$$
On cross multiplying, we get
$$3x^2+6kx-j=0$$
Here $$a = 3, b = 6k, c = -j$$
On substituting the values, we get
$$=$$ $$\dfrac{-6k\pm\sqrt{(6k)^2-4\times 3\times -j}}{2\times 3}$$
$$=$$ $$\dfrac{-6k\pm\sqrt{36k^2+12 j}}{6}$$
$$=$$ $$\dfrac{-6k\pm\sqrt{12(3k^2+j)}}{6}$$
$$=$$ $$\dfrac{-6k\pm4\sqrt{3(3k^2+j)}}{6}$$
Take $$2$$ as common, we get
$$=$$ $$-3k\pm\dfrac{\sqrt{3(3k^2+j)}}{3}$$
Therefore, $$x =-k\pm \dfrac{\sqrt{3(3k^2+j)}}{3}$$
What are the real factors of $$x^2 + 4$$?
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$$(x^2+2)$$ and $$(x^2-2)$$
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$$(x+2)$$ and $$(x-2)n$$
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Does not exist
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$$(x^2 + 2)$$ and $$(x + 2)$$
Explanation
Real factors of the equation does not exist as it doesn't have real roots
$$x^{2}=-4$$
$$\Rightarrow x=+2i$$ or $$- 2i$$
Which of the following is an approximate of a zero of the equation $$x^2 - 3x = 7$$?
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-4.54
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-1.54
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1.54
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3.54
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5.54
Explanation
$$x^2-3x-7=0$$
$$x=\dfrac{3\pm \sqrt{9+28}}{2}$$ (using $$x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$$ for, $$ax^2+bx+c=0$$)
$$=\dfrac{3\pm 6.08}{2}$$
$$x=4.54$$ & $$x=-1.54$$
$$x=4.54$$ not in options
$$\therefore x=-1.54$$(Approximate zero).
If A and B are whole numbers such that $$9A^{2} = 12A + 96$$ and $$B^{2} = 2B + 3$$, find the value of $$5A + 7B$$.
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$$31$$
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$$37$$
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$$41$$
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$$43$$
Explanation
A,B are whole numbers
$$9{ A }^{ 2 }=12A+96$$ & $${ B }^{ 2 }=2B+3$$
$$9{ A }^{ 2 }-12A-96=0$$ & $${ B }^{ 2 }-2B-3=0$$
$$3{ A }^{ 2 }-4A-32=0$$
$$3{ A }^{ 2 }-12A+8A-32=0$$ & $${ B }^{ 2 }-3B+B-3=0$$
$$(3A+8)(A-4)=0$$ & $$(B-3)(B+1)=0$$
$$\therefore A=4,B=3$$ ($$\because A,B$$ are whole numbers $$\because (A\neq \cfrac { -8 }{ 3 }) $$ &$$(B\neq -1$$)
$$5A+7B=5(4)+7(3)$$
$$=41$$
Given that '$$x$$' is real then the solution set of the equation $$\sqrt { x-1 } +\sqrt { x+1 } =1$$.
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No solution
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Unique solution
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$$2$$ solutions
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None of these
Explanation
$$\sqrt{x-1}+\sqrt{x+1}=1$$ As $$x > 1$$
$$\because x-1$$ can't be negative.
$$\Rightarrow \sqrt{x+1} > 1$$ $$0\sqrt{x-1}$$ is possitve
$$\therefore $$ there sum is $$ > 1$$ always . hence no solution.
Let $$a$$ be the solution of the equation $$4x^2-12x+9=16$$, then the value of $$10a-15$$ is
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20
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25
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30
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35
Explanation
$$4x^2 - 12x + 9 = 16$$
$$\Rightarrow 4x^2 - 12x - 7 = 0$$
$$\Rightarrow x = \cfrac{12 \pm \sqrt{144 + 112}}{8}$$
$$\Rightarrow x = \cfrac{12 \pm 16}{8}$$
$$\Rightarrow x = -0.5, 3.5$$
Substituting the value of a in $$10a - 15$$, we get $$35 - 15 = 20$$ and $$5 - 15 = -10$$
If $$x = 7 + 4\sqrt {3}$$ and $$xy = 1$$, what is the value of $$\dfrac {1}{x^{2}} + \dfrac {1}{y^{2}}$$?
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$$64$$
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$$134$$
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$$194$$
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$$\dfrac {1}{49}$$
Explanation
$$x=7+4\sqrt{3}$$ & $$xy=1$$
$$\Rightarrow y=\dfrac{1}{7+4\sqrt{3}}=\dfrac{7-4\sqrt{3}}{49-48}=7-4\sqrt{3}$$
$$\dfrac{1}{x^{2}}+\dfrac{1}{y^{2}}=(\dfrac{1}{x}+\dfrac{1}{y})^{2}-2 \Rightarrow \dfrac{1}{x^{2}}+\dfrac{1}{y^{2}}=(\dfrac{1}{7+4\sqrt{3}}+\dfrac{1}{7-4\sqrt{3}})^{2}-2$$
$$=(\dfrac{14}{49-48})^{2}-2=196-2$$
$$=\boxed{194}$$
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
Discrimant (D) is applied only to quadratic equations, but the equation
$$ \sqrt{x^{2}+x+1} = 2 $$ is not an quadratic equation,
Hence, first convert the equation into
quadratic by squaring it on both sides,
$$ \sqrt{x^{2}+x+1} = 2 $$
$$ \Rightarrow x^{2}+x+1 = 4 $$
$$ \Rightarrow x^{2}+x-3 = 0 $$
$$ \therefore D = b^{2}-4ac $$
$$ D = 1-4\times 1\times (-3) $$
$$ \boxed{D = 13} $$
If $$f(x)=x^2+4x+a$$ is a perfect square function then calculate the value of $$a$$.
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$$3$$
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$$4$$
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$$2$$
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$$6$$
Explanation
Since f(x) is perfect square, then both the roots are equal.
$$\therefore$$ D=0
$$16-4a=0$$
$$a=4$$
Identify which of the following is/are a quadratic polynomial function:
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$$f(x)=(x+1)^3-(x+2)^3$$
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$$g(x)=\dfrac{x^4}{x^2} $$
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$$h(x)=(x+1)^2-(x+2)^2$$
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All of these
Explanation
Option 1) $$f(x) =(x+1)^3-(x+2)^3$$
$$=x^3+1+3x(x+1)-{x^3+8+6x(x+2)}$$
$$=x^3+1+3x^2+3x-x^3-8-6x^2-12x$$
The power of $$x^3$$ cancels out and results in to quadratic equation.
Option 2) g(x) =$$x^2$$
Option 3) $$x^2+1+2x-x^2-4-8x$$
The power of $$x^2$$ cancels out which doesn't result in to quadratic equation.
The roots of the quadratic equation $$(a+b-2c)x^2-(2a-b-c)x+(a-2b+c)=0$$ are-
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$$(a+b+c)$$ and $$(a-b-c)$$
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$$\dfrac 12$$ and $$a-2b+c$$
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$$a-2b+c$$ and $$\dfrac 1{(a+b-2c)}$$
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None of the above.
Explanation
Clearly we see that x=1 satisfies the equation
$$\Rightarrow $$x=1 is a root of the that equation
Let $$\alpha $$ be other root
Product of roots =$$1×\alpha =\dfrac{a-2b+c}{a+b-2c}$$
other root $$=\alpha =\dfrac {a-2b+c}{a+b-2c}$$
Sum of the roots of the equation $$(x+3)^2-4|x+3|+3=0$$ is-
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$$4$$
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$$12$$
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$$-12$$
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$$-4$$
Explanation
$${ (x+3) }^{ 2 }-4|x+3|+3=0\\ let\quad |x+3|=t\\ \therefore t\ge 0\\ { t }^{ 2 }-4t+3=0\\ (t-1)(t-3)=0\\ \therefore t=1\ or\ 3\\ when\quad t=1\\ |x+3|=1\\ x+3=1\quad or\quad x+3=-1\\ x=-2\quad or\quad x=-4\\ when\quad t=3\\ |x+3|=3\quad or\quad x+3=-3\\ x=0\quad or\quad x=-6\\ sum\quad of\quad all\quad roots\quad =\quad (-2)+(-4)+(0)+(-6)\\ =-12$$
The roots of the equation , $$(x^2+1)^2=x(3x^2+4x+3)$$, are given by-
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$$2-\sqrt 3$$
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$$\dfrac {-1+i\sqrt 3}2$$
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$$2+\sqrt 3$$
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$$\dfrac {-1-i\sqrt 3}2$$
Explanation
$${ ({ x }^{ 2 }+1) }^{ 2 }=x(3{ x }^{ 2 }+4x+3)\\ { x }^{ 4 }+2{ x }^{ 2 }+1=3{ x }^{ 3 }+4{ x }^{ 2 }+3x\\ { x }^{ 4 }-{ 3x }^{ 3 }-2{ x }^{ 2 }-3x+1=0\\ \therefore ({ x }^{ 2 }-4x+1=0)({ x }^{ 2 }+x+1)=0\\ { x }^{ 2 }-4x+1=0\quad or\quad { x }^{ 2 }+x+1=0\\ \therefore x=\cfrac { 4\pm \sqrt { 16-4 } }{ 2 } \quad \\ x=\cfrac { 4\pm \sqrt [ 2 ]{ 3 } }{ 2 } \\ x=2\pm \sqrt { 3 } \\ \therefore x=(2+\sqrt { 3 } )\ or\ (2-\sqrt { 3 } )\\ x=\cfrac { -1\pm \sqrt { 1-4 } }{ 2 } \\ x=\cfrac { -1\pm i\sqrt { 3 } }{ 2 } \\ x=\cfrac { -1+i\sqrt { 3 } }{ 2 }\ or\ \cfrac { -1-i\sqrt { 3 } }{ 2 } $$
If $$\alpha,\beta$$ are roots of the equation $$2x^2-35x+2=0$$ then the value of $$(2\alpha-35)^3(2\beta-35)^3$$ is:
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$$1$$
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$$8$$
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$$4$$
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$$64$$
Explanation
We will substitute $$\alpha, \beta $$ in the above equation
$$\Rightarrow 2\alpha - 35+\dfrac {2}{\alpha} =0$$
$$\Rightarrow 2\alpha - 35=\dfrac {-2}{\alpha} $$
Similarly for beta
$$\Rightarrow (2\alpha - 35)^{3}(2\beta - 35)^{3}=\dfrac {-8}{\alpha ^{3}}×\dfrac {-8}{\beta^{3}}=\dfrac {64}{1}=64$$
$$A$$ and $$B$$ solve an equation $$x^2+px+q=0$$. In solving $$A$$ commits a mistake in reading $$p$$ and finds the root $$2$$ and $$6$$ and $$B$$ commits a mistake in reading $$q$$ and finds the roots $$2$$ and $$-9$$. Find the correct roots.
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$$2$$
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$$-3$$
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$$-4$$
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$$-2$$
Explanation
$${ x }^{ 2 }+px+q=0$$
As A copied q correct so he will get product of roots correct
As B copied P correct so he will get sum of roots correct
$$sum\quad =\quad 2-9=-7\\ product\quad =\quad 2\times6=12\\ \therefore equation\\ { x }^{ 2 }+7x+12=0\\ (x+4)(x+3)=0\\ \therefore x=-3\quad or\quad -4$$
The minimum value of the expression $$4x^2+2x+1$$ is-
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$$1$$
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$$\dfrac{1}{3}$$
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$$\dfrac{1}{2}$$
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$$\dfrac{3}{4}$$
Explanation
For a general quadratic equation $$ { ax }^{ 2 }+bx+c=0,\quad where\quad a>0\quad the\quad minimum\quad value\quad of\quad expression\quad occurs\quad at\quad \cfrac { -D }{ 4a } \\ \therefore { 4x }^{ 2 }+2x+1\\ minimum\quad value\quad =\quad \cfrac { -({ 2 }^{ 2 }-4.4.1) }{ 4.4 } \\ =\cfrac { 12 }{ 16 } =\cfrac { 3 }{ 4 } $$
The number of values of $$a$$ for which $$(a^2-3a+2)x^2+(a^2-5a+6)x+a^2-4=0$$ is an identity in $$x$$ is-
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$$0$$
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$$2$$
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$$1$$
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$$3$$
Explanation
Equation is a identity if all its coefficients are equal to zero simultaneously
$$\Rightarrow a^{2}-3a+2=0$$
$$\Rightarrow a=2,1$$
$$\Rightarrow a^{2}-5a+6=0$$
$$\Rightarrow a^{2}-4=0$$
$$\Rightarrow a=\pm 2$$
$$\Rightarrow$$coefficients are equal to zero simultaneously when $$a=2$$
Therefore only one value of 'a' the equation is identity in $$x$$
If $$b\in F'$$ then the roots of the equation $$\left( 2+b \right) { x }^{ 2 }+(3+b)x+(4+b)=0\quad $$ is
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real and imaginary
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real and equal
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imagenary
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cannot predicted
Explanation
$$(2+b)x^{2}+(3+b)x+(4+b)=0$$
$$D=(3+b)^{2}-4(2+b)(4+b)=b^{2}+6{b}+9-4{b^{2}}-24{b}-32=-3{b^{2}}-18{b}-23=-3(b^{2}+6{b}+9)+4$$
$$=-3(b+3)^{2}+9<0$$
So roots are imaginary
The number of integral value of k such that the given quadratic equation has imaginary roots are?
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$$0$$
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$$\sqrt{2/3}$$
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$$2$$
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$$3$$
If $$a, b,c$$ are distinct and the roots of $$(b-c)x^{2} +(c-a)x +(a-b)=0$$ are equal, then $$a,b,c$$ are in:
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Arithmetic Progression
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Geometric progression
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Harmonic progression
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Arithmetico-Geometric progression
Explanation
$$f\left(x\right)=\left(b-c\right)x^{2}+\left(c-a\right)x+\left(a-b\right)=0$$
Given that the roots are equal.
$$\Rightarrow\Delta=0$$
Now,
$$\Delta=b^{2}-4ac=0$$
$$\Rightarrow\left(c-a\right)^{2}-4\left(b-c\right)\left(a-b\right)=0$$
$$\Rightarrow{c}^{2}-2ac+a^{2}+4b^{2}+4ac-4bc-4ab=0$$
$$\Rightarrow\left(a-2b+c\right)^{2}=0$$
$$\Rightarrow\left(a-2b+c\right)=0$$ $$\Rightarrow2b=a+c$$
$$\Rightarrow{a},b, c$$ are in A.P.
The discriminant value of equation $$5{x}^{2}-6x+1=0$$ is ...............
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$$16$$
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$$\sqrt { 56 } $$
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$$4$$
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$$56$$
Explanation
The given equation is $$5{x}^{2}-6x+1=0$$ ........... $$(i)$$
General form of quadratic equation is given by
$$ax^2+bx+c=0$$ ....... $$(ii)$$
Comparing equation $$(i)$$ with $$(ii)$$
we have,
$$a=5,b=-6,c=1$$
The discriminant $$D$$ is given by
$$D={b}^{2}-4ac={(-6)}^{2}-4(5)(1)=36-20=16$$
$$\therefore$$ Discriminant value is $$D=16$$
............... is true for discriminate of quadratic equation $$x^2 + x + 1 = 0$$.
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D = 0
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D < 0
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D > 0
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D is a perfect square
Explanation
The polynomial is $$x^2 + x + 1 = 0$$.
Comparing with $$ax^2 + bx + c = 0$$, we get
$$a = 1, b = 1, c = 1$$
$$D = b^2 - 4ac = 1 - 4(1) (1) = -3 < 0$$
Solve the following quadratic equation by completing the square: $$\dfrac{5x+7}{x-1}=3x+2$$
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$$\left \{ -1, 3 \right \}$$
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$$\left \{ 1, 3 \right \}$$
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$$\left \{ -1, -3 \right \}$$
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None of these
Explanation
Given $$\dfrac{5x+7}{x-1}=3x+2$$
$$\Rightarrow$$ $$5x+7=(3x+2)(x-1)$$
$$\Rightarrow$$ $$5x+7=3x^2-3x+2x-2$$
$$\Rightarrow$$ $$3x^2-6x-9=0$$
$$\Rightarrow$$ $$x^2-2x-3=0$$
$$\Rightarrow$$ Here, $$a=1,\,b=-2,\,c=-3$$
$$\Rightarrow$$ $$x^2-2x=3$$
Now, $$\left(\dfrac{b}{2}\right)^2=\left (\dfrac{-2}{2} \right )^2=1$$
Adding $$1$$ on both sides,
$$\Rightarrow$$ $$x^2-2x+1=3+1$$
$$\Rightarrow$$ $$(x-1)^2=4$$
Taking square root on both sides
$$\Rightarrow$$ $$x-1=\pm2$$
$$\Rightarrow$$ $$x-1=2$$ and $$x-1=-2$$
$$\therefore$$ $$x=3$$ and $$x=-1$$
Let $$\alpha$$ be the root of the equation $$25\cos^{2}\theta + 5\cos \theta - 12 = C$$, where $$\dfrac {\pi}{2} < \alpha < \pi$$.
What is $$\tan \alpha$$ equal to?
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$$-\dfrac {3}{4}$$
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$$\dfrac {3}{4}$$
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$$-\dfrac {4}{3}$$
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$$-\dfrac {4}{5}$$
Explanation
Given equation is $$25\cos^2\theta+5\cos\theta-12=0$$
$$\Rightarrow 25\cos^2\theta+20\cos\theta-15\cos\theta-12=0$$
$$\Rightarrow 5\cos\theta(5\cos\theta+4)-3(5\cos\theta+4)=0$$
$$\Rightarrow (5\cos\theta-3)(5\cos\theta+4)=0$$
So, $$\cos\theta=\dfrac{3}{5}$$ or $$\cos\theta=\dfrac{-4}{5}$$
Since $$\alpha$$ is the root of the equation and it lies between $$\dfrac{\pi}{2}<\alpha<\pi\Rightarrow \cos \alpha\ $$ is '-ve'.
Thus
$$\alpha=\cos^{-1}\left (\dfrac{-4}{5}\right)$$
Since it lies in the range
$$\dfrac{\pi}{2}<\alpha<\pi$$, so $$\tan \alpha$$ is '-ve'
$$\tan\alpha=\tan\left [\cos^{-1}\left (\dfrac{-4}{5}\right)\right]$$
$$=\tan \left [\tan^{-1}\left (-\dfrac{3}{4}\right)\right]$$
$$=-\dfrac{3}{4}$$
If the roots of the equation $$x^{2} + px + c = 0$$ are $$(2, -2)$$ and the roots of the equation $$x^{2} + bx + q = 0$$ are $$(-1, -2)$$, then the roots of the equation $$x^{2} + bx + c = 0$$ are
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$$-3, -2$$
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$$-3, 2$$
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$$1, -4$$
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$$-5, 1$$
Explanation
Given, roots of the equation $$x^{2} + px + c = 0$$ are $$2$$ and $$-2$$.
$$\therefore -p = 2 - 2 \Rightarrow p = 0$$
and $$(2)\cdot (-2) = c\Rightarrow c = -4$$
Again, roots of the equation $$x^{2} + bx + q = 0$$ are $$-1$$ and $$-2$$.
$$\therefore -b = -1 -1 \Rightarrow b = 3$$
and $$(1-) \cdot (-2) = q\Rightarrow q = 2$$
$$\therefore x^{2} + bc + c \equiv x^{3} + 3x - 4 =0$$
$$\Rightarrow (x - 1)(x + 4) = 0$$
So, the required roots are $$1$$ and $$-4$$.
Sum of the roots of the equation $${ \left| x-3 \right| }^{ 2 }+\left| x-3 \right| -2=0$$ is
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$$2$$
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$$4$$
0%
$$6$$
0%
$$16$$
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$$-2$$
Explanation
Given equation is $${ \left| x-3 \right| }^{ 2 }+\left| x-3 \right| -2=0$$
$$\Rightarrow { \left| x-3 \right| }^{ 2 }+2\left| x-3 \right| -\left| x-3 \right| -2=0$$
$$\Rightarrow \left| x-3 \right| \left( \left| x-3 \right| +2 \right) -1\left( \left| x-3 \right| +2 \right) =0$$
$$\Rightarrow \left( \left| x-3 \right| +2 \right) \left( \left| x-3 \right| -1 \right) =0$$
Therefore, $$ \left| x-3 \right| =-2,1$$
But $$\left| x-3 \right| \neq -2$$
Hence, $$\left| x-3 \right| =1$$
$$\Rightarrow x-3=\pm 1$$
$$\Rightarrow x=3\pm 1$$
$$\Rightarrow x=4, 2$$
Now, sum of the roots $$=4+2=6$$.
The equation $${ e }^{ \sin { x } }-{ e }^{ -\sin { x } }-4=0$$ has
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No solution
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Two solutions
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Three solutions
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None of these
Explanation
Given, $${ e }^{ \sin { x } }-{ e }^{ -\sin { x } }-4=0$$
multiply throughout by $$e^{sinx}$$
$$\Rightarrow { e }^{ 2\sin { x } }-4{ e }^{ \sin { x } }-1=0$$
this is quadratc equation in $$e^{sinx}$$
converting from exponential to logarithmic form.
$${ e }^{ \sin { x } }=\dfrac { 4\pm \sqrt { 16+4 } }{ 2 } =2\pm \sqrt { 5 }$$
$$\Rightarrow \sin { x } =\log { \left( 2+\sqrt { 5 } \right) } $$ [$$ \because \log { \left( 2-\sqrt { 5 } \right) }$$ is not defined ]
Since, $$2+\sqrt { 5 } > e \Rightarrow \log { \left( 2+\sqrt { 5 } \right) } > 1$$
$$\Rightarrow \sin { x } > 1$$, which is not possible.
Hence, no solution exist.
The root of the equation $$2(1+i)x^2-4(2-i)x-5-3i=0$$, where $$i=\sqrt{-1}$$, which has eater modulus is
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$$(3-5i)/2$$
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$$(5-3i)/2$$
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$$(3+i)/2$$
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$$(1+3i)/2$$
Which of the following equations, is not a quadratic equation?
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$$4x^{2} - 7x + 3 = 0$$
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$$3x^{2} - 4x + 1 = 0$$
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$$2x - 7 = 0$$
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$$4x^{2} - 3 = 0$$
Explanation
Alternate $$(A) : 4x^{2} - 7x + 3 = 0$$
The maximum exponent of variable $$x$$ is $$2$$. So it is a quadratic equation.
Alternate $$(B) : 3x^{2} - 4x + 1 = 0$$
The maximum exponent of variable $$x$$ is $$2$$. So it is a quadratic equation.
Alternate $$(C) : 2x - 7 = 0$$
The maximum exponent of variable $$x$$ is $$1$$. So it is not a quadratic equation.
The discriminant of quadratic equation $$3x^{2} - 4x - 1 = 0$$ is _______.
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$$0$$
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$$4$$
0%
$$12$$
0%
$$28$$
Explanation
Quadratic equation: $$3x^{2} - 4x - 1 = 0$$
Comparing the given equation with $$ax^{2} + bx + c = 0$$
we have $$a = 3, b = -4, c = -1$$
Now, discriminant
$$D = b^{2} - 4ac$$
$$= (-4)^{2} - 4(3) (-1)$$
$$= 16 + 12$$
$$= 28$$.
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