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CBSE Questions for Class 10 Maths Quadratic Equations Quiz 9 - MCQExams.com
CBSE
Class 10 Maths
Quadratic Equations
Quiz 9
Solve the quadratic equation by using quadratic formula $$9x^2-3( a+b) x+ ab=0$$
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$$x=\dfrac {a}{3}, \dfrac {b}{3}$$
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$$x=\dfrac {-a}{3}, \dfrac {-b}{3}$$
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$$x=\dfrac {-a}{3}, \dfrac {b}{3}$$
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$$x=\dfrac {a}{3}, \dfrac {-b}{3}$$
Explanation
$$\Rightarrow$$ Here, $$a=9,\,b=-3(a+b),\,c=ab$$
$$\Rightarrow$$ $$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$$
$$=\dfrac{-[-3(a+b)]\pm\sqrt{[-3(a+b)]^2-4(9)(ab)}}{2(9)}$$
$$=\dfrac{3a+3b\pm\sqrt{9(a^2+b^2+2ab)-36ab}}{18}$$
$$=\dfrac{3a+3b\pm\sqrt{9a^2+9b^2+18ab-36ab}}{18}$$
$$=\dfrac{3a+3b\pm\sqrt{9a^2+9b^2-18ab}}{18}$$
$$\dfrac{3a+3b\pm\sqrt{9(a^2+b^2-2ab)}}{18}$$
$$=\dfrac{3a+3b\pm\sqrt{9(a-b)^2}}{18}$$
$$=\dfrac{3a+3b\pm3(a-b)}{18}$$
$$=\dfrac{3a+3b\pm 3a-3b}{18}$$
$$x=\dfrac{3a+3b+3a-3b}{18}$$ and $$x=\dfrac{3a+3b-3a+3b}{18}$$
$$x=\dfrac{6a}{18}$$ and $$x=\dfrac{6b}{18}$$
$$\therefore$$ $$x=\dfrac{a}{3},\,\dfrac{b}{3}$$
The solution for $$2x^2-9x+7=0$$ is
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$$2,3$$
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$$1,\dfrac 94$$
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$$\dfrac 14,\dfrac {13}4$$
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$$1,\dfrac {13}4$$
Explanation
Given Quadratic equation is
$$2x^2-9x+7=0$$
Comparing it with the general form of quadratic equation $$ax^2+bx+c=0$$
we write, $$a=2,b=-9,c=7$$
Quadratic formula is $$x=\dfrac{-b\pm \sqrt {b^2-4ac}}{2a}$$
$$\implies x=\dfrac{-(-9)\pm\sqrt {(-9)^2-4(2)(7)}}{2(2)}$$
$$\implies x=\dfrac{9\pm\sqrt {81-56}}{4}$$
$$\implies x=\dfrac{9\pm\sqrt {25}}{4}$$
$$\implies x=\dfrac{9\pm 5}{4}$$
$$\implies x=\dfrac{9 +5}{4} \quad x=\dfrac{9-5}4$$
$$\implies x=1,\dfrac{7}2$$
Find the roots of the following equations, if they exist, by the completing the square:
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$$2x^{2}-7x+3=0$$
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$$2x^{2}+x-4=0$$
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$$4x^{2}+\sqrt4{3x}+3=0$$
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$$2x^{2}+x+4=0$$
Explanation
$$so\quad the\quad root\quad of\quad quadratic\quad equation\quad are\quad$$ $$X=3\quad and\quad X=\dfrac { 1 }{ 2 } $$
The root of $$x^{2}+kx+k=0$$ are real and equal , find $$k$$.
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$$0$$
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$$4$$
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$$0$$ or $$4$$
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$$2$$
Explanation
The roots of $$x^2 + kx + k = 0$$ are real and equal
i.e., $$b^2 - 4ac = 0$$
$$\implies k^2 - 4(1)k = 0$$
$$\implies k^2 - 4k = 0$$
$$\implies k(k - 4) = 0$$
$$\therefore k = 0 $$ or $$4$$
Find the root of the following quadratic equation, if they exist, by the method of completing the square.
$$2x^2+x-4=0$$.
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$$ \dfrac{-1-{\sqrt 33}}{4} $$
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$$ \dfrac{-1-{\sqrt 33}}{44} $$
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$$ \dfrac{-1+{\sqrt 33}}{44} $$
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$$ \dfrac{1-{\sqrt 33}}{4} $$
If both roots of quadratic equation $$(\alpha +1)x^{2}-2(1+3\alpha )x+1+8\alpha =0$$ are real and distinict, the $$\alpha$$ be-
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-2
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1
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2
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3
Explanation
$$\left( \alpha + 1 \right) {x}^{2} - 2 \left( 1 + 3 \alpha \right) x + 1 + 8 \alpha = 0$$
Here,
$$a = \alpha + 1$$
$$b = - 2 \left( 1 + 3 \alpha \right)$$
$$c = 1 + 8 \alpha$$
If both roots are real and distinct,
$$D > 0$$
$$\Rightarrow {b}^{2} - 4ac > 0$$
$$\Rightarrow {\left( - 2 \left( 1 + 3 \alpha \right) \right)}^{2} - 4 \left( 1 + \alpha \right) \left( 1 + 8 \alpha \right) > 0$$
$$\Rightarrow 4 \left( 1 + 9 {\alpha}^{2} + 6 \alpha \right) - 4 \left( 1 + 9 \alpha + 8 {\alpha}^{2} \right) > 0$$
$$\Rightarrow 4 \left( 1 + 9 {\alpha}^{2} + 6 \alpha - 1 - 9 \alpha - 8 {\alpha}^{2} \right) > 0$$
$$\Rightarrow {\alpha}^{2} - 3 \alpha > 0$$
$$\Rightarrow \alpha \left( \alpha - 3 \right) > 0$$
$$\Rightarrow \alpha \in \left( - \infty, 0 \right) \cup \left( 3, \infty \right)$$
Hence among the given values, $$\alpha$$ will be $$-2$$.
Hence the correct answer is $$\left( A \right) -2$$.
Identify the standard form of the given equation$$(4x-5)(4x+5)=0$$.
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$$16x^2-25=0$$
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$$16x^3-25=0$$
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$$16x^2-2=0$$
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$$6x^2-25=0$$
Explanation
$$ (4x+5)(4x-5) = 0$$
$$ 4x(4x-5)+5(4x-5) = 0$$
$$ 16x^2-20x+20x-25 = 0$$
$$ 16x^2-25 =0 $$ which is in standard form of quadratic equation
If $$x^{2}-5x+1=0$$, then $$\dfrac {x^{10}+1}{x^{5}}$$ has the value
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$$2524$$
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$$2525$$
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$$2424$$
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$$2010$$
Explanation
$$x^2-5x+1=0$$
$$\therefore x^2+1=5x$$
$$\therefore x+\dfrac{1}{x}=5$$
$$\therefore x^2+\dfrac{1}{x^2}=25-2=23$$
$$\therefore x^3+\dfrac{1}{x^3}=\left(x+\dfrac{1}{x}\right)^3-3\left(x+\dfrac{1}{x}\right)$$
$$\therefore x^3+\dfrac{1}{x^3}=125-15$$
$$\therefore x^3+\dfrac{1}{x^3}=110$$
Now $$\left(x^3+\dfrac{1}{x^2}\right)\left(x^3+\dfrac{1}{x^3}\right)=56\times 23$$
$$\therefore x^5+x+\dfrac{1}{x}+\dfrac{1}{x^5}=110\times 23$$
$$\therefore x^5+\dfrac{1}{x^5}=2530-5$$
$$\therefore x^5+\dfrac{1}{x^5}=2525$$.
If $$a,b,c,x$$ are real numbers and $$\left( { a }^{ 2 }+{ b }^{ 2 } \right) { x }^{ 2 }-2b\left( a+c \right) x+\left( { b }^{ 2 }+{ c }^{ 2 } \right) =0$$ has real & equal roots, then $$a,b,c$$ are in
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$$A.P$$
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$$G.P$$
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$$H.P$$
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$$None of these$$
Explanation
Since $$(a^2+b^2)x^2-2b(a+c)x+(b^2+c^2)=0$$ has real & equal roots.
$$\therefore$$ $$4b^2(a^2+c^2+2ac)-4(a^2+b^2)(b^2+c^2)=0$$
$$4b^2a^2+4b^2c^2+8acb^2-4a^2b^2-4a^2c^2-4b^2-4b^2c^2=0$$
$$8acb^2-4a^2c^2+4b^4=0$$
$$(2ac-2b^2)^2=0$$
$$\boxed{ac=b^2}$$
The discriminant of the quadratic equation $$5 x ^ { 2 } - 6 x + 1 = 0$$ 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 quadratic equation is $$5x^2-6x+1=0$$ is in the form of $$ax^2+bx+c$$
by comparing $$a=5, b=-6, c=1$$
Now, Discriminant $$D=b^2-4ac$$
$$D=36-4\times 5 \times 1$$
$$D=36-20=16$$
Let $${ \lambda }_{ 1 }$$ and $${\lambda}_{2}$$ be two values of $${\lambda}$$ for which the expression $${x}^{2}+(2-\lambda)x+\lambda-{3}$$ becomes a perfect square. The value of $$(\lambda_{1}^{2}+\lambda_{2}^{2})$$ equals
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$$32$$
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$$25$$
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$$50$$
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$$100$$
$$\begin{array} { l } { \text { If } \alpha , \beta \text { are the roots of } a x ^ { 2 } + b x + c \text { and } \alpha + h } , { \beta + h \text { are the roots of } p x ^ { 2 } + q x + r = 0 } \\{ \text { and } D _ { 1 } \text { , } D _ { 2 } \text { are the respective } } { \text { discriminants of these equations, } } \\ { \text { then } D _ { 1 } : D _ { 2 } = } \end{array}$$
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$$\cfrac{a ^ { 2 } }{ p ^ { 2 }}$$
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$$\cfrac{b ^ { 2 }} { q ^ { 2 }}$$
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$$\cfrac{c ^ { 2 } }{r ^ { 2 }}$$
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$$1$$
Explanation
Let $$A=\alpha+h$$ and $$B=\beta+h$$
Then,$$A-B=\left(\alpha+h\right)-\left(\beta+h\right)$$
$$\Rightarrow\,A-B=\alpha-\beta$$
$$\Rightarrow\,{\left(A-B\right)}^{2}={\left(\alpha-\beta\right)}^{2}$$
$$\Rightarrow\,{\left(A+B\right)}^{2}-4AB={\left(\alpha+\beta\right)}^{2}-4\alpha\beta$$
$$\Rightarrow\,\dfrac{{b}^{2}}{{a}^{2}}-\dfrac{4c}{a}=\dfrac{{q}^{2}}{{p}^{2}}-\dfrac{4r}{p}$$
$$\Rightarrow\,\dfrac{{b}^{2}-4ac}{{a}^{2}}=\dfrac{{q}^{2}-4rp}{{p}^{2}}$$
$$\Rightarrow\,\dfrac{{D}_{1}}{{a}^{2}}=\dfrac{{D}_{2}}{{p}^{2}}$$
$$\Rightarrow\,\dfrac{{D}_{1}}{{D}_{2}}=\dfrac{{a}^{2}}{{p}^{2}}$$
Hence $${D}_{1}:{D}_{2}={a}^{2}:{p}^{2}=\dfrac{{a}^{2}}{{p}^{2}}$$
Find the root of the quadratic equation $${x^2} + 2\sqrt {2x} + 6 = 0$$ by using the quadratic formula
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$$x = \sqrt 2 \pm 2i$$
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$$x = - \sqrt 2 \pm 2i$$
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$$x = - \sqrt 4 \pm 2i$$
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$$x = - \sqrt 2 \pm 4i$$
Explanation
According to the given quadratic equation,
$$a=1,b=2\sqrt2$$ and $$c=6$$
So, by quadratic formula,
$$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}=\dfrac{-2\sqrt2\pm\sqrt{(2\sqrt2)^2-4\times1\times6}}{2\times1}$$
$$=\dfrac{-2\sqrt2\pm\sqrt{8-24}}{2}=\dfrac{-2\sqrt2\pm\sqrt{-16}}{2}=\dfrac{-2\sqrt2\pm4i}{2}$$
Thus, $$x=-\sqrt2\pm2i$$
Let f (x) = $$x^{2}+\lambda x+\mu cosx,\lambda $$ is +ve integer $$\mu $$ is a real number. The number of ordered pairs ($$\lambda ,\mu $$) for which f (x) = 0 and f (f(c)) = 0 have same set of real roots.
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0
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1
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2
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3
If $$\dfrac{\sqrt{1296}}{x}=\dfrac{x}{2.25}$$ then value $$x$$ is:
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$$0.9$$
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$$0.09$$
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$$9$$
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$$1.9$$
Explanation
we know that, $$\sqrt {1296} = 36$$
$$\dfrac {36}{x} = \dfrac{x}{2.25}$$
$$x ^ 2 = 36 \times 2.25$$
$$x ^ 2 = 81$$
$$x = \pm \sqrt {81}$$
$$x =\pm 9$$
Find the value of $$x:$$
$$x^2-4x+4=0$$
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$$2$$
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$$1$$
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$$0$$
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None of the above
Explanation
Given equation is:
$$x^2-4x+4=0$$
It can also be written as:
$$x^2-2(2)(x)+(2)^2=0\dots (1)$$
Above equation is of the form of following identity:
$$(a-b)^2=a^2-2ab+b^2$$
Hence, the equation $$(1)$$ can be written as follows:
$$(x-2)^2=0$$ or
$$(x-2)(x-2)=0$$
$$x=2,2$$
If $$k> 0$$ and the product of roots of the equations $$x^{2}-3kx+2e^{\log k}-1=0$$ is $$7$$ , then find the sum of the roots-
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$$6$$
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$$4$$
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$$3$$
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$$-12$$
Explanation
Given: $$ { x^{ 2 } }-3kx+2.{ e^{ 2\log k } }-1=0 $$
$$\Rightarrow { x^{ 2 } }-3kx+2{ e^{ \log { k^{ 2 } } } }-1=0$$
$$ \Rightarrow { x^{ 2 } }-3kx+2{ e^{ \log { k^{ 2 } } } }-1=0 $$
$$\Rightarrow { x^{ 2 } }-3kx+\left( { 2{ k^{ 2 } }-1 } \right) =0 $$
Let $$ \alpha$$
and $$\beta$$ be the roots
Hence $$\alpha +\beta =3k $$
and $$ \alpha \beta =2{ k^{ 2 } }-1=7$$
$$ Rightarrow 2{ k^{ 2 } }=8=4 $$
$$\Rightarrow k=\pm 2 $$
$$\Rightarrow k=2$$ So, $$k>0$$
Now, $$\alpha +\beta =3k$$
$$=3\left( 2 \right) =6$$
If $$x ^ { 2 } - 5 x + 1 = 0$$ then $$\frac { x ^ { 10 } + 1 } { x ^ { 5 } }$$ is equal to:-
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2424
0%
3232
0%
2525
0%
None of these
$$(x+1)^2+(y-1)^2 +(x-5)^2+(y-1)^2$$ has the value equal to
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$$16$$
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$$25$$
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$$36$$
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$$49$$
The ratio of the roots of the equation $$ { ax }^{ 2 }+bx+c=0$$ is same as the ratio of the roots of equation $$ { px }^{ 2 }+qx+r=0$$. If $$ { D }_{ 1 }$$ and $$ { D }_{ 2 }$$ are the discriminants of $$ { ax }^{ 2 }+bx+c=0$$ and $$ { px }^{ 2 }+qx+r=0$$ respectively, then $$ { D }_{ 1 }:{ D }_{ 2 }=$$
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$$ \dfrac { { b }^{ 2 } }{ { q }^{ 2 } } $$
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$$ \dfrac { b }{ q } $$
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$$ \dfrac { b }{ { q }^{ 2 } } $$
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none of these
Explanation
$$\textbf{Step 1: Using quadratic formula,}$$
$$\boldsymbol{x=\dfrac{-b\pm\sqrt{D}}{2a}}$$
$$ax^2+bx+c=0$$
$$x=\dfrac{-b\pm\sqrt{D_1}}{2a}$$ $$[\text{ Using Quadratic Formula }]$$
$$px^2+qx+r=0$$
$$x=\dfrac{-q\pm\sqrt{D_2}}{2a}$$
$$[\text{ Using Quadratic Formula }]$$
$$\text{As per given question, We have }$$
$$\dfrac{-b+\sqrt{D_1}}{-b-\sqrt{D_1}}=\dfrac{-q+\sqrt{D_2}}{-q-\sqrt{D_2}}$$
$$\textbf{Step 2: Using Componendo and Dividendo, we get}$$
$$\dfrac{-\sqrt{D_1}}{b}=\dfrac{-\sqrt{D_2}}{q}$$
$$\text{On squaring both side}$$
$$\dfrac{D_1}{b^2}=\dfrac{D_2}{q^2}$$
$$\Rightarrow \dfrac{D_1}{D_2}=\dfrac{b^2}{q^2}$$
Which of the following is a quadratic equation ?
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$$6x^2 = 20 -x^3$$
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$$x^2-\left ( \dfrac{1}{x^2} \right ) $$ = $$\dfrac{7}{2}$$
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$$\dfrac{3}{x}$$ $$ = 4$$ $$x^2$$
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$$5x^2 + 7 = 3x$$
Explanation
For $$(A)$$
$$6x^2=20-x^3$$
$$\Rightarrow x^3+6x^2-20=0$$
In this, the maximum power of $$x$$ is $$3$$.
So, it is not a quadratic equation.
For $$(B)$$
$$x^2-\left(\dfrac{1}{x^2}\right)=\dfrac{7}{2}$$
$$\Rightarrow 2x^4-2=7x^2$$
$$\Rightarrow 2x^4-7x^2-2=0$$
In this, the maximum power of $$x$$ is $$4$$.
So, it is not a quadratic equation.
For $$(C)$$
$$\dfrac{3}{x}=4x^2$$
$$\Rightarrow 3=4x^3$$
In this, the maximum power of $$x$$ is $$3$$.
So, it is not a quadratic equation.
For $$(D)$$
$$5x^2+7=3x$$
$$\Rightarrow 5x^2-3x+7=0$$
In this, the maximum power of $$x$$ is $$2$$.
So, it is a quadratic equation.
Hence, $$Op-D$$ is correct.
Solve the quadratic equation by completing the square method: $${x}^{2}+8x-9=0$$
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$$-9,-1$$
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$$2,1$$
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$$-9,1$$
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$$9,1$$
Explanation
$$x^2+8x-9=0$$
$$\Rightarrow x^2+2\times \left( 4\right) \times x +16-25=0$$
$$\left( x+4\right)^2=25$$
$$\left( x+4\right)^2=5^2$$
$$x+4=\pm 5$$
$$\Rightarrow x=1$$ and $$-9$$
Hence, the answer is $$1,-9.$$
Discriminant is ................. for the equation $$5x-6=-\frac{1}{x}$$
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-56
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16
0%
-16
0%
0
If $$(1+m^2)x^2-2(1+3m)x+(1+8m)=0$$. Then numbers of value(s) of m for which this equation has no solution.
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$$3$$
0%
infinitely many
0%
$$2$$
0%
$$1$$
Explanation
We know that $$D=b^2-4ac$$
$$D=4(1+3m)^2-4(1+m^2)(1+8m)$$
$$=4(1+9m^2+6m-(1+8m+m^2+8m^3))$$
$$=4(8m^2-2m-8m^3)$$
$$=-8(4m^3-4m^2+m)$$
$$=-8m(4m^2-4m+1)$$
$$=-8m(2m-1)^2 < 0$$.........$$\because (2m-1)^2 \, is \, always +ve$$
$$=m>0$$
Hence infinitely many values.
Quadratic equation among the following is:
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$${ x }^{ 2 }+1={ x }^{ 2 }+2x$$
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$$x\left( x+1 \right) =x\left( x-3 \right) $$
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$${ x }^{ 2 }-2x=3\left( 5x+2 \right) $$
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$${ x }^{ 2 }+2x+3={ x }^{ 2 }+5x$$
For what value of $$c$$, the root of $$ (c-2) x^{2}+2(c-2) x+2=0$$
are not real
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$$( 1,2)$$
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$$( 2,3)$$
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$$( 3,4)$$
0%
$$(2,4)$$
Explanation
Given the quadratic equation is
$$ (c-2) x^{2}+2(c-2) x+2=0$$.
Now the roots of the equation will be not real if the discriminant of the equation is less than $$0$$.
Then,
$$\{2(c-2)\}^2-4.2.(c-2)<0$$
or, $$4\{(c-2)(c-4)\}<0$$
or, $$(c-2)(c-4)<0$$
or, $$2<c<4$$.
So $$c\in (2,4)$$.
Which constant must be added and subtracted to solve the quadratic equation $$9x^2 + \dfrac{3}{4}x - \sqrt{2} = 0$$ by the method of completing the square ?
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$$\dfrac{1}{8}$$
0%
$$\dfrac{1}{64}$$
0%
$$\dfrac{1}{4}$$
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$$\dfrac{9}{64}$$
Explanation
Given equation is $$9x^2 + \dfrac{3}{4}x - \sqrt{2} = 0$$
$$\Rightarrow (3x)^2 + 2\left(\dfrac{1}{8}\right)(3x) - \sqrt{2} = 0$$
Hence, to make it a perfect square we must add and subtract $$\left(\dfrac{1}{8}\right)^2=\dfrac1{64}$$
If $$\alpha$$ and $$\beta$$ are roots of the equation $$x^2 + x\sin \theta - 2 \sin \theta = 0 $$ then $$\dfrac{\alpha^{12} + \beta^{12}}{(\alpha^{-12} + \beta^{-12})(\alpha - \beta)^{24}}$$ is equal to
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$$\dfrac{2^{24}}{(8 + \sin \theta)^{12}}$$
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$$\dfrac{2^{12}}{(8 + \sin \theta)^{12}}$$
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$$\dfrac{2^{12}}{(8 - \sin \theta)^{12}}$$
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$$\dfrac{1}{2^{24}}$$
Explanation
$$x^2 + x\sin \theta - 2 \sin \theta = 0$$, has roots $$\alpha$$ and $$\beta$$
$$\dfrac{\alpha^{12}+\beta^{12}}{(\alpha^{-12}+\beta^{-12})(\alpha - \beta)^{24}} = \dfrac{\alpha^{12} . \beta^{12}}{(\alpha - \beta)^{24}} = \dfrac{(\alpha \beta)^{12}}{(\alpha - \beta)^{24}}$$
$$ = \dfrac{(-2\sin \theta)^{12}}{(\sqrt{\sin^2 \theta + 8 \sin \theta})^{24}} = \dfrac{2^{12}}{(8 + \sin \theta)^{12}}$$
If the discriminant of $$3 x ^ { 2 } - 2 x + K = 0$$ is zero then $$\mathrm { K } =$$ _____
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$$3$$
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$$\dfrac { 1 } { 3 }$$
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$$-3$$
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$$\dfrac { -1 } { 3 }$$
Which constant must be added and subtracted to solve the Quadratic equation $$9 x ^ { 2 } + \dfrac { 3 } { 4 } x - \sqrt { 2 }$$ by the method of
completion the square.
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0%
$$\dfrac { 1 } { 8 }$$
0%
$$\dfrac { 1 } { 4 }$$
0%
$$\dfrac { 1 } { 64 }$$
0%
$$\dfrac { 9 } { 64 }$$
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