Quadratic Equations Test

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Quadratic Equations Test - 39

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Weekly Quiz Competition

Question 1
1 / -0

Find the roots of the following equations, if they exist, by the completing the square:

A
$$2x^{2}-7x+3=0$$

B
$$2x^{2}+x-4=0$$

C
$$4x^{2}+\sqrt4{3x}+3=0$$

D
$$2x^{2}+x+4=0$$

Solution

$$so\quad the\quad root\quad of\quad quadratic\quad equation\quad are\quad$$ $$X=3\quad and\quad X=\dfrac { 1 }{ 2 } $$
Question 2
1 / -0

Find the root of the following quadratic equation, if they exist, by the method of completing the square.
$$2x^2+x-4=0$$.

A
$$ \dfrac{-1-{\sqrt 33}}{4} $$

B
$$ \dfrac{-1-{\sqrt 33}}{44} $$

C
$$ \dfrac{-1+{\sqrt 33}}{44} $$

D
$$ \dfrac{1-{\sqrt 33}}{4} $$
Question 3
1 / -0

The root of $$x^{2}+kx+k=0$$ are real and equal , find $$k$$.

A
$$0$$

B
$$4$$

C
$$0$$ or $$4$$

D
$$2$$

Solution

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 $$4$$
Question 4
1 / -0

Identify the standard form of the given equation$$(4x-5)(4x+5)=0$$.

A
$$16x^2-25=0$$

B
$$16x^3-25=0$$

C
$$16x^2-2=0$$

D
$$6x^2-25=0$$

Solution

$$ (4x+5)(4x-5) = 0$$
$$ 4x(4x-5)+5(4x-5) = 0$$
$$ 16x^2-20x+20x-25 = 0$$
$$ 16x^2-25 =0 $$ which is in standard form of quadratic equation
Question 5
1 / -0

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-

A
-2

B
1

C
2

D
3

Solution

$$\left( \alpha + 1 \right) {x}^{2} - 2 \left( 1 + 3 \alpha \right) x + 1 + 8 \alpha = 0$$
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 $$\left( A \right) -2$$.
Question 6
1 / -0

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

A
$$A.P$$

B
$$G.P$$

C
$$H.P$$

D
$$None of these$$

Solution

Since $$(a^2+b^2)x^2-2b(a+c)x+(b^2+c^2)=0$$ has real & equal roots.
$$\therefore$$ $$4b^2(a^2+c^2+2ac)-4(a^2+b^2)(b^2+c^2)=0$$
$$4b^2a^2+4b^2c^2+8acb^2-4a^2b^2-4a^2c^2-4b^2-4b^2c^2=0$$
$$8acb^2-4a^2c^2+4b^4=0$$
$$(2ac-2b^2)^2=0$$
$$\boxed{ac=b^2}$$
Question 7
1 / -0

Roots of the equation $$\sqrt {\dfrac {x}{1-x}}+\sqrt {\dfrac {1-x}{x}}=2\dfrac {1}{6}$$ are

A
$$\dfrac {3}{2},\dfrac {2}{3}$$

B
$$\dfrac {9}{13},\dfrac {4}{13}$$

C
$$\dfrac {1}{3},\dfrac {3}{1}$$

D
$$\dfrac {1}{13},\dfrac {13}{1}$$

Solution

$$\sqrt{\dfrac{x}{1-x}}+\sqrt{\dfrac{1-x}{x}}=2\dfrac{1}{6}$$

$$\Rightarrow$$ $$\sqrt{\dfrac{x}{1-x}}+\sqrt{\dfrac{1-x}{x}}=\dfrac{13}{6}$$

$$\Rightarrow$$ $$\left(\sqrt{\dfrac{x}{1-x}}+\sqrt{\dfrac{1-x}{x}}\right)^2=\left(\dfrac{13}{6}\right)^2$$

$$\Rightarrow$$ $$\left(\sqrt{\dfrac{x}{1-x}}\right)^2+\left(\sqrt{\dfrac{1-x}{x}}\right)^2+2\times \sqrt{\dfrac{x}{1-x}}\times \sqrt{\dfrac{1-x}{x}}=\dfrac{169}{36}$$

$$\Rightarrow$$ $$\dfrac{x}{1-x}+\dfrac{1-x}{x}+2=\dfrac{169}{36}$$

$$\Rightarrow$$ $$\dfrac{x^2+1+x^2-2x}{x-x^2}+\dfrac{2x-2x^2}{x-x^2}=\dfrac{169}{36}$$

$$\Rightarrow$$ $$36(1)=169(x-x^2)$$
$$\Rightarrow$$ $$36=169x-169x^2$$
$$\Rightarrow$$ $$169x^2-169x+36=0$$
$$\Rightarrow$$ $$169x^2-117x-52x+36=0$$
$$\Rightarrow$$ $$13x(13x-9)-4(13x-9)=0$$
$$\Rightarrow$$ $$(13x-9)(13x-4)=0$$
$$\Rightarrow$$ $$13x-9=0$$ and $$13x-4=0$$
$$\therefore$$ $$x=\dfrac{9}{13}$$ and $$x=\dfrac{4}{13}$$
Question 8
1 / -0

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

A
$$32$$

B
$$25$$

C
$$50$$

D
$$100$$

Solution
Question 9
1 / -0

The discriminant of the quadratic equation $$5 x ^ { 2 } - 6 x + 1 = 0$$ is

A
$$16$$

B
$$\sqrt { 56 }$$

C
$$4$$

D
$$56$$

Solution

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 $$D=b^2-4ac$$

$$D=36-4\times 5 \times 1$$

$$D=36-20=16$$
Question 10
1 / -0

$$\begin{array} { l } { \text { If } \alpha , \beta \text { are the roots of } a x ^ { 2 } + b x + c \text { and } \alpha + h } , { \beta + h \text { are the roots of } p x ^ { 2 } + q x + r = 0 } \\{ \text { and } D _ { 1 } \text { , } D _ { 2 } \text { are the respective } } { \text { discriminants of these equations, } } \\ { \text { then } D _ { 1 } : D _ { 2 } = } \end{array}$$

A
$$\cfrac{a ^ { 2 } }{ p ^ { 2 }}$$

B
$$\cfrac{b ^ { 2 }} { q ^ { 2 }}$$

C
$$\cfrac{c ^ { 2 } }{r ^ { 2 }}$$

D
$$1$$

Solution

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 $${D}_{1}:{D}_{2}={a}^{2}:{p}^{2}=\dfrac{{a}^{2}}{{p}^{2}}$$

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Re-Attempt Weekly Quiz Competition

Quadratic Equations Test - 39

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Quadratic Equations Test - 39

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SHARING IS CARING

Question 1

$$2x^{2}-7x+3=0$$

$$2x^{2}+x-4=0$$

$$4x^{2}+\sqrt4{3x}+3=0$$

$$2x^{2}+x+4=0$$

Solution

Question 2

$$ \dfrac{-1-{\sqrt 33}}{4} $$

$$ \dfrac{-1-{\sqrt 33}}{44} $$

$$ \dfrac{-1+{\sqrt 33}}{44} $$

$$ \dfrac{1-{\sqrt 33}}{4} $$

Question 3

$$0$$

$$4$$

$$0$$ or $$4$$

$$2$$

Solution

Question 4

$$16x^2-25=0$$

$$16x^3-25=0$$

$$16x^2-2=0$$

$$6x^2-25=0$$

Solution

Question 5

-2

1

2

3

Solution

Question 6

$$A.P$$

$$G.P$$

$$H.P$$

$$None of these$$

Solution

Question 7

$$\dfrac {3}{2},\dfrac {2}{3}$$

$$\dfrac {9}{13},\dfrac {4}{13}$$

$$\dfrac {1}{3},\dfrac {3}{1}$$

$$\dfrac {1}{13},\dfrac {13}{1}$$

Solution

Question 8

$$32$$

$$25$$

$$50$$

$$100$$

Solution

Question 9

$$16$$

$$\sqrt { 56 }$$

$$4$$

$$56$$

Solution

Question 10

$$\cfrac{a ^ { 2 } }{ p ^ { 2 }}$$

$$\cfrac{b ^ { 2 }} { q ^ { 2 }}$$

$$\cfrac{c ^ { 2 } }{r ^ { 2 }}$$

$$1$$

Solution

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