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CBSE Questions for Class 10 Maths Quadratic Equations Quiz 4 - MCQExams.com
CBSE
Class 10 Maths
Quadratic Equations
Quiz 4
Which of the following is non-quadratic polynomial.
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$$x ^ { 2 } + 2$$
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$$x ^ { 2 } + 3 x$$
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$$4 x ^ { 4 } + 3 y$$
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$$x ^ { 2 }$$
Explanation
$$x^2+2, x^2+3x$$ and $$x^2$$ are all quadratic polynomials. But $$4x^4+3y$$ is not a quadratic polynomial because this polynomial has a degree $$4$$.
The discriminent of graph is
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$$b ^ { 2 } - 4 a c = 0$$
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$$b ^ { 2 } - 4 a c > 0$$
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$$b ^ { 2 } - 4 a c < 0$$
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$$b ^ { 2 } +4 a c < 0$$
Explanation
Since the graph is touching the $$x-$$ axis at only one point
so the quadratic equation has equal roots
$$\implies \mathrm{discriminant}=0$$
$$\implies b^2-4 {a}{c}=0$$
Which of the following must be added to $$x^2-6x+5$$ to make it a perfect square ?
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$$3$$
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$$4$$
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$$5$$
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$$6$$
Explanation
We've,
$$x^2-6x+5$$
$$=x^2-2.x.3+3^2-4$$
$$=(x-3)^2-4$$.
So to make it perfect square i.e. $$(x-3)^2$$, we are to add $$4$$ to add.
The discriminant of the quadratic equation $$ ax^2+bx+c= 0$$ i
s
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$$\Delta = b^2 + 4ac$$
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$$\Delta = b^2 - 4ac$$
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$$\Delta =4abc$$
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$$\Delta = b^2 \times 4ac$$
Explanation
The discriminant of the quadratic equation $$ax^2 + bx + c = 0$$ is
$$\Delta = b^2 - 4ac$$
Values of $$k$$ for which the quadratic equation $$2x^2+ kx + k = 0$$ has equal roots.
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$$4, 8$$
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$$0, 4$$
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$$4,-8$$
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$$0,8$$
Explanation
For equal roots, discriminant D of equation $$2x^2+kx+k=0$$, should be zero.
$$D =k^2-4(2)(k)=0$$
$$k^2-8k=0$$
$$k(k-8)=0$$
$$k=0, 8$$
Which of the following is a quadratic equation?
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$$x^3 + 2x + 1 = x^2 + 3$$
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$$-2x^2=(5-x)\left ( 2x-\cfrac{2}{5}\right )$$
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$$(k+1)x^2+\cfrac{3}{2}x=7,$$where $$k=-1$$
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$$x^3 x^2 = (x)^3$$
Which constant must be added and subtracted to solve the quadratic equation $$\displaystyle 9x^2 +\frac{3}{4}x+ 2 = 0$$ by the method of completing the square?
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$$\displaystyle \frac{1}{64}$$
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$$\displaystyle \frac{1}{576}$$
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$$\displaystyle \frac{1}{144}$$
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$$\displaystyle \frac{127}{64}$$
Explanation
$$9x^2 + \dfrac 34x +2 = 0$$
Dividing by $$9$$ on both sides we get,
$$x^2 + \dfrac{x}{12} + \dfrac 29 = 0$$
$$x^2 + \dfrac{x}{12} = -\dfrac 29$$
Third term $$= \left(\dfrac 12 \times\text{coefficient of }x\right)^2$$
$$= \left(\dfrac 12 \times\dfrac {1}{12}\right)^2$$ = $$\left(\dfrac {1}{24}\right)^2 =\dfrac{1}{576}$$
Adding $$\dfrac {1}{576}$$ on both sides we get,
$$ \therefore x^2 + \dfrac{x}{12} + \dfrac {1}{576}$$ = $$\dfrac{1}{576} - \dfrac 29$$
$$\left(x + \dfrac{1}{24}\right)^2 =$$ $$\dfrac{1}{576} - \dfrac 29$$
Hence, option $$B$$.
Which of the following is not a quadratic equation :
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$$(x-2)^{2}+1=2x-3$$
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$$x(x+1)+8=(x+2)(x-2)$$
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$$x(2x+3)=x^{2}+1$$
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$$(x+2)^{3}=x^{3}-4$$
Explanation
$$x(x+1)+8=(x+2)(x-2)$$
$$\therefore x^2+x+8=x^2-4$$
$$\therefore x+12=0$$
The highest power of $$x$$ in this equation is $$1$$.
So, this is not a quadratic equation.
The answer is option (B)
(Note that, in option $$D$$,
$$(x+2)^3 = (x^3-4)$$ also reduces to quadratic, since the $$x^3$$ terms cancel out.)
If the expression $$(a-2)x^{2}+2(2a-3)x+(5a-6)$$ is positive for all real values of $$x$$, then
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$$a$$ can be any real number
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$$a>1$$
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$$a>3$$
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$$a=3$$
Explanation
The expression is always positive that means the discriminant of the equation is always less than zero .
So, $$4(2a-3)^2-4(a-2)(5a-6)<0$$
$$(16a^2+36-48a)-(20a^2-64a+48)<0$$
$$(-4a^2+16a-12)<0$$
$$(-4)(a^2-4a+3)<0$$
$$(a-3)(a-1)>0$$
So, the expression will be positive for $$a>3$$ and $$a<1$$
Option C is correct .
lf the roots of $${p}{x}^{2}+2{q}{x}+{r}=0$$ and $$qx^{2}-2\sqrt{pr}x+q=0$$ are simultaneously real, then
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$$ p=q$$ ; $$r\neq 0$$
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$$ 2q=\sqrt{pr}$$
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$$ pr=q^{2}$$
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$$ {p}{r}={q}$$
Explanation
Since roots are real,
$$\Rightarrow (2q)^{2}-4pr\geq 0$$ and $$(2\sqrt{pr})^{2}-4q^{2}\geq 0$$
$$4q^{2}\geq 4pr$$ and $$4pr\geq 4q^{2}$$
To hold above two equations simultenously
$$\therefore q^{2}=pr$$
If $$a > 0$$, then the expression $$ax^{2}+bx+c$$ is positive for all values of $$x$$ provided
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$$b^{2}-4ac> 0$$
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$$b^{2}-4ac< 0$$
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$$b^{2}-4ac= 0$$
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$$b^{2}-ac< 0$$
Explanation
Consider the equation,
$$ax^{2}+bx+c$$
Now if $$b^{2}-4ac<0$$, then it does not have any real roots and hence it will not intersect the $$x$$ axis at any point.
If $$a>0$$ then $$y=ax^{2}+bx+c$$ will lie completely above $$x$$ axis and
If $$a<0$$ then $$y=ax^{2}+bx+c$$ will lie completely below $$x$$ axis.
It is given that the above equation is always greater than zero,
Or
$$ax^{2}+bx+c>0$$ for $$\epsilon R$$.
This can only occur is
$$b^{2}-4ac<0$$ and $$a>0$$.
If $$a=0$$, then the equation $$\displaystyle \frac{x-a-1}{x-a}=a +1-\displaystyle \frac{1}{x-a}$$ has
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one root.
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two roots.
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many roots.
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no roots.
Explanation
$$\dfrac{x-a-1}{x-a}=a+1-\dfrac{1}{x-a}$$
$$1-\dfrac{1}{x-a}=a+1-\dfrac{1}{x-a}$$
Since, given $$a=0$$, equation holds for all $$x$$ except $$x=a$$
$$\therefore $$ The equation has many roots.
Which of the following is not a quadratic equation?
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$$2(x + 1)^2 = 4x^2 + 2x + 1$$
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$$2x + x^2 = 2x^2 + 5$$
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$$( \sqrt{2} \times \sqrt{3}x)^2+ x^2 = 3x^2 + 5x$$
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$$(x^2 + 2x)^2 = x^5 + 3 + 4x^3$$
Explanation
A. $$2(x+1)^{2}=4x^{2}+2x+1$$
$$2(x^{2}+2x+1)=4x^{2}+2x+1$$
$$2x^{2}+4x+2=4x^{2}+2x+1$$
$$2x^{2}+4x+2-4x^{2}-2x-1=0$$
$$-2x^{2}+2x+1=0$$
is a quadratic equation
B. $$2x+x^2=2x^2+5$$
$$-x^2+2x=5$$
is a quadratic equation
C. $$[\sqrt{2}x\sqrt{3}]^2+x^2=3x^2+5x$$
$$(6x^2)+x^2-3x^2-5x=0$$
$$4x^2-5x=0$$
is a quadratic equation
D. $$(x^2+2x)^2=x^5+4x^3+3$$
As degree of $$x$$ is $$5$$, it is not quadratic.
The equation $$(a+2){x}^{2}+(a-3)x=2a-1,a\neq -2$$ has rational roots for
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all rational values of $$a$$ except $$a=-2$$
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all real values of $$a$$ except $$a=-2$$
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rational values of $$a> \displaystyle \frac{1}{2}$$
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none of these
If $$14$$ is the maximum of $$-\lambda x^{2} + \lambda x + 8$$, then the value of $$\lambda$$ is
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$$24$$
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$$\sqrt[6]{3}$$
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$$\sqrt[-6]{3}$$
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$$- 12$$
Explanation
$$-\lambda x^{2} + \lambda x + 8$$
$$=-\lambda(x^2 - x - \cfrac{8}{\lambda})$$
$$=- \lambda({(x - \cfrac{1}{2})^2 - \cfrac{1}{4} - \cfrac{8}{\lambda})})$$
$$=-\lambda(x - \cfrac{1}{2})^2) + \cfrac{\lambda}{4} + 8$$
Maximum value of the expression is: $$8 + \cfrac{\lambda}{4}$$
$$8 + \cfrac{\lambda}{4} = 14$$
$$\lambda= 24$$
If equation $$x^2-(2+m) x+1(m^2-4m+4)=0 $$ has coincident roots, then :
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$$m=0$$
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$$m = 6$$
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$$m=2$$
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$$m \displaystyle = \frac{2}{3}$$
Explanation
Given: $$x^2-(2+m)x+1(m^2-4m+4)=0$$ has coincident roots
Therefore $$b^2 - 4ac=0$$
here $$a=1, \ b=-(2+m),\ c=(m^2-4m+4)$$
$$\Rightarrow [-(2+m)]^2-4\times 1\times (m^2-4m+4)=0$$
$$\Rightarrow 4+m^2+4m-4m^2-16m-16=0$$
$$\Rightarrow -3m^2+20m-12=0$$
$$\Rightarrow 3m^2 - 20m +12=0$$
$$\Rightarrow 3m^2 - 2m -18m+12=0$$
$$\Rightarrow m(3m - 2)-6(3m -2)=0$$
$$\Rightarrow (m-6)(3m-2)=0$$
$$\therefore\ \ m = \cfrac{2}{3} , 6$$
If the roots of the equation $$x^{2} + a^{2} = 8x + 6a$$ are real, then:
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$$a \epsilon \left [ 2,8 \right ]$$
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$$ a \epsilon \left [ -2,8 \right ]$$
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$$a \epsilon (2,8)$$
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$$ a \epsilon (-2,8)$$
Explanation
The given equation can be written as
$$ x^{2} - 8x + a^{2} -6a = 0$$
Since the roots of the above equation are real,
$$\therefore B^{2} -4AC \geq 0 \Rightarrow 64-4(a^{2}-6a)\geq 0$$
$$\Rightarrow a^{2} -6a -16 \leq 0$$
$$\Rightarrow (a+2)(a-8) \leq 0$$
$$\Rightarrow a\ \epsilon\ \left [ -2,8 \right ]$$
Which of the following is not a quadratic equation?
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$$2(x-1)^2=4x^2-2x+1$$
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$$(x^2+1)^2=x^2+3x+9$$
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$$(x^2+2x)^2=x^4+3+4x^3$$
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$$x^2+9=3x^2-5x$$
Explanation
It is not a quadratic equation because, here maximum power of x is 4.
Find the roots of the following quadratic equation
$$z^2\,+\,6z\,-\,8\,=\,0$$
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$$z\,=\,-3\,\pm\,\sqrt{19}$$
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$$z\,=\,-3\,\pm\,\sqrt{17}$$
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$$z\,=\,-2\,\pm\,\sqrt{17}$$
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$$z\,=\,-3\,\pm\,\sqrt{15}$$
Explanation
The roots of the quadratic equation $$ax^2+bx+c=0$$ are
$$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$$
So, the roots of $$z^2+6z-8=0$$ are
$$z=\dfrac{-6\pm\sqrt{6^2-4(1)(-8)}}{2}$$
$$\therefore z=\dfrac{-6\pm\sqrt{68}}{2}$$
$$\therefore z=\dfrac{-6\pm2\sqrt{17}}{2}$$
$$\therefore z=-3\pm \sqrt{17}$$
Find the discriminant for the given quadratic equation:
$$4x^2\,-\,kx\,+\,2\,=\,0$$
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$$k^2\,-\,21$$
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$$k^2\,-\,24$$
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$$k^2\,-\,28$$
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$$k^2\,-\,32$$
Explanation
Given equation is,
$$4x^2 - kx + 2 = 0 $$
We know, $$D = b^2 - 4ac $$
$$a=4, b=-k, c=2$$
$$D = (-k)^2 - 4(4)(2)$$
$$\therefore D = k^2 - 32$$
If both 'a' and 'b' belong to the set $$\left \{ 1, 2, 3, 4 \right \}$$, then the number of equations of the form $$ax^{2}+bx+1=0$$ having real roots is:
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$$10$$
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$$7$$
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$$6$$
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$$12$$
Explanation
For a quadratic equation to have a real roots, discriminant must be greater than or equal to zero
$$\Rightarrow b^{2}-4a\geq0$$
Since, $$a,b \in \{1,2,3,4\}$$
(1) When $$a=1$$
$$\Rightarrow b^{2}\geq 4$$
$$\Rightarrow b=2,3,4$$
(2) When $$a=2$$
$$\Rightarrow b^{2}\geq 8$$
$$\Rightarrow b=3,4$$
(3) When $$a=3$$
$$\Rightarrow b^{2}\geq 12$$
$$\Rightarrow b=4$$
(3) When $$a=4$$
$$\Rightarrow b^{2}\geq 16$$
$$\Rightarrow b=4$$
Therefore, total $$7$$ possibilities are there.
Hence the answer is $$7$$.
If the equation $$4x^{2} + x(p + 1) + 1 = 0$$ has exactly two equal roots, then one of the values of $$p$$ is
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$$5$$
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$$-3$$
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$$0$$
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$$3$$
Explanation
For the equation $$4x^{2} + x(p + 1) + 1 = 0$$ to have two equal roots, the condition is Discriminant $$=0$$
$$b^{2}-4ac = 0$$
$$(p+1)^{2}-4\times 4\times 1 = 0$$
$$p^{2}+2p+1-16 = 0$$
$$p^{2}+2p-15 = 0$$
$$(p+5)(p-3) = 0$$
$$\therefore p = -5$$ or $$p = 3$$
Hence, one of the values of $$p$$ is $$3$$.
Option D is correct.
The value of $$x^{2} - 6x + 13$$ can never be less than
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$$4$$
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$$5$$
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$$4.5$$
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$$7$$
Explanation
We know that a perfect square can never be less than $$0$$.
$$\because x^{2}-6x+13 = x^{2}-6x+9+4$$
$$= (x-3)^{2}+4$$
because $$(x-3)^{2}$$ can never be less than zero
$$\therefore$$ least value of $$x^{2}-6x+13\ is \ 4$$
If $$\displaystyle ax^{2}+bx+1= 0, a \in R, b\in R,$$ does not have distinct real roots, then the maximum value of $$b^2$$ is
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$$-4a$$
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$$4a$$
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$$2a$$
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$$-2a$$
Explanation
$$The\quad equation\quad a{ x }^{ 2 }-bx+1=0\quad has\quad no\quad distinct\quad
real\quad roots,\\ This\quad implies\quad D=b^2-4ac\le 0,\\ b^2-4a\le 0 \\ b^2\le 4a$$.
If $$\displaystyle x^{2}-bx+c=0$$ has equal integral roots, then
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$$b$$ and $$c$$ are integers
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$$b$$ and $$c$$ are even integers
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$$b$$ is an even integer and $$ c$$ is $$a$$ perfect square of $$a$$ positive integer
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none of these
Explanation
The roots of this equation will be
$$\alpha =\dfrac{b\pm \sqrt{b^{2}-4c}}{2}$$
Both roots are equal so
$$\dfrac{b+\sqrt{b^{2}-4c}}{2}=\dfrac{b-\sqrt{b^{2}-4c}}{2}$$
$$b=\pm 2\sqrt{c}$$
Therefore, $$c$$ should be perfect square integer and $$b$$ is even number.
Hence, options 'A' and 'C' are correct.
If $$\displaystyle ax^{2}+bx+6=0$$ does not have two distinct real roots, where $$a\in R,b\in R$$, then the least value of $$3a+b$$ is
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$$4$$
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$$-1$$
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$$1$$
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$$-2$$
Explanation
$$ax^{ 2 }+bx+6=0$$
Substitute $$b=k-3a$$ to obtain $$ax^{ 2 }+\left( k-3a \right) x+6=0$$
Since, the equation does not have real distinct roots.
Therefore, $$D={ \left( k-3a \right) }^{ 2 }-24a\le 0$$
$$\Rightarrow 9a^{ 2 }-6a\left( 4+k \right) +{ k }^{ 2 }\le 0$$
Above quadratic equation has real roots.
Therefore, $$D=36\left[ { \left( 4+k \right) }^{ 2 }-{ k }^{ 2 } \right] \ge 0$$
$$\Rightarrow k\ge -2$$
$$\Rightarrow 3a+b\ge -2$$.
Ans: D
The number of values of $$k$$ for which $$\displaystyle \left \{x^{2}-(k-2)x+k^{2}\right\}+ \left \{x^{2}+kx+(2k-1)\right \}$$ is a perfect square is/are
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$$1$$
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$$2$$
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$$0$$
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none of these
Explanation
Consider $$p\left( x \right) =\left\{ x^{ 2 }-(k-2)x+k^{ 2 } \right\} +\left\{ x^{ 2 }+kx+(2k-1) \right\} $$
$$\Rightarrow p\left( x \right) =2{ x }^{ 2 }+2x+{ k }^{ 2 }+2k-1$$
$$p\left( x \right) $$ is perfect square when roots of the equation $$p\left( x \right) =0$$ are equal
Therefore, $$D={ b }^{ 2 }-4ac=4-8\left( { k }^{ 2 }+2k-1 \right) =0$$
$$\Rightarrow 2{ k }^{ 2 }-4k-3=0$$
Therefore, number of values of $$k$$ are 2.
Ans: B
If at least one of the equations $$ { x^{ 2 } }+px+q=0$$ , $$ { x^{ 2 } }+rx+s=0 $$ has real roots, then
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$$qs =(p+r)$$
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$$pr =(q+s)$$
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$$pr =2(q+s)$$
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None of these.
Explanation
Given at least one of the equation $$x^{ 2 }+px+q=0$$ , $$x^{ 2 }+rx+s=0 $$ has real roots.
Let $$D_1$$ and $$D_2$$ be the discriminant values.
At least one of the equation has real roots when $$D_1+D_2 \geq 0$$.
$$\Rightarrow p^2-4q+r^2-4s \geq 0$$
$$\Rightarrow (p-r)^2+2pr-4(q+s) \geq 0$$
Above inequality holds only if $$2pr-4(q+s)=0$$.
$$\therefore pr=2(q+s)$$.
Hence, option C.
If $$x$$ is real, the expression $$\displaystyle \frac { \left( x-a \right) \left( x-c \right) }{ \left( x-b \right) } $$ is capable of assuming all values if
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$$a > b > c$$
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$$a < b < c$$
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$$a < b > c$$
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Cannot say
Explanation
$$y=\dfrac { \left( x-a \right) \left( x-c \right) }{ \left( x-b \right) } $$
$$(x-b)y={ x }^{ 2 }-(a+c)x+ac$$
$${ x }^{ 2 }-(a+c+y)x+ac+by=0$$
$$D\ge 0$$
$${ (a+c+y) }^{ 2 }-4(ac+by)\ge 0$$
$${ y }^{ 2 }+{ a }^{ 2 }+{ c }^{ 2 }+2ac+2cy+2ay-4ac-4by\ge 0$$
$${ y }^{ 2 }+(2a+2c-4b)y+{ (a-c) }^{ 2 }\ge 0$$
So $$D\le 0$$
So $$4{ (a+c-2b) }^{ 2 }-4{ (a-c) }^{ 2 }\le 0$$
$$(a+c-2b+a-c)(a+c-2b-a+c)\le 0$$
$$(a-b)(c-b)\le 0$$
$$a < b < c$$
$$a > b > c$$
For the quadratic equation $$\displaystyle 2x^{2}+6\sqrt{2}x+1=0$$
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roots are rational
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if one root is $$p+\sqrt{q}$$, then the other is $$-p+\sqrt{q}$$
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roots are irrational
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if one root is $$p+\sqrt{q}$$, then the other is $$p-\sqrt{q}$$
Explanation
Here,
$$2x^{ 2 }+6\sqrt { 2x } +1=0$$
$$\displaystyle \Rightarrow x=\frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a } =\frac { -6\sqrt { 2 } \pm \sqrt {64} }{ 4 } =\frac { -3\sqrt { 2 } \pm 4 }{ 2 } $$
Therefore, roots are irrational.
If $$p=-\dfrac{3\sqrt{2}}{2}$$ and $$q={4}$$
One root is $$p+\sqrt{q}$$, then the other root is
$$p-\sqrt{q}$$
Ans: C,D
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