Quadratic Equations Test

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Quadratic Equations Test - 37

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Weekly Quiz Competition

Question 1
1 / -0

Find the roots of the following quadratic equation (if they exist) by the method of completing the square.
$$x^2 \, - \, (\sqrt{2} \, + \, 1) \, x \, + \, \sqrt{2} \, = \, 0$$

A
$$exist, \sqrt2$$, $$1$$

B
$$exist, -\sqrt2$$, $$1$$

C
$$exist, -\sqrt2$$, $$-1$$

D
$$\text{does not exist}$$

Solution

$$x^2-(\sqrt{2}+1)x+\sqrt{2}=0$$
Here, $$a=1,\,b=-(\sqrt{2}+1),\,c=\sqrt{2}$$
$$\Rightarrow$$ $$x^2-(\sqrt{2}+1)x=-\sqrt{2}$$

Now, $$\left(\dfrac{\sqrt{2}+1}{2}\right)^2=\dfrac{2+2\sqrt{2}+1}{4}$$

Adding $$\dfrac{2+2\sqrt{2}+1}{4}$$, on both sides,

$$\Rightarrow$$ $$x^2-(\sqrt{2}+1)x+\dfrac{2+2\sqrt{2}+1}{4}=-\sqrt{2}+\dfrac{2+2\sqrt{2}+1}{4}$$

$$\Rightarrow$$ $$\left[x-\left(\dfrac{\sqrt{2}+1}{2}\right)\right]^2=\dfrac{2-2\sqrt{2}+1}{4}$$

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

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

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

$$\therefore$$ $$x=\sqrt{2}$$ and $$x=1$$
Question 2
1 / -0

Using the identity $$\left( {x + a} \right)\left( {x + b} \right),$$ the value of $$\left( {2a - 1} \right)\left( {2a + 2} \right)$$

A
$$4{a^2} + 2a - 2$$

B
$$-4{a^2} + 2a + 2$$

C
$$2\left( {2{a^2} + a - 1} \right)$$

D
$$4{a^2} + 2a + 2$$

Solution

$$\left(2a-1\right)\left(2a+2\right)$$
$$={4a}^{2}-2a+4a-2$$
$$={4a}^{2}+2a-2$$
Question 3
1 / -0

Which of the following statements is TRUE/CORRECT about Quadratic Equations? A quadratic equation is _____

A
An expression with a single variable and degree $$2$$.

B
An equation with a single variable and degree $$2$$.

C
An equation with degree $$2$$ only.

D
An equation with single variable only.

Solution

The definition of a Quadratic Equation states that an equation with one variable and degree $$2$$ is a Quadratic Equation. Only option $$b$$ satisfies the two basic conditions of a quadratic equation i.e. one/single variable and degree $$2$$. Hence $$b$$ is the correct answer.
Question 4
1 / -0

Which of the following methods is used to derive the Standard Quadratic Formula for the Quadratic Equation $$ax^2+bx+c=0$$?

A
Factorisation Method

B
Completing Square Method

C
Hit and Trial Method

D
All the above.

Solution

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.
Question 5
1 / -0

If ($$\alpha $$, $$\beta $$) are the roots of $$a{x^2} + bx + c = 0$$ ,$$\left( {a \ne 0} \right)$$ and ($$\alpha + \delta $$ , $$\beta + \delta $$) are the roots of $$A{x^2} + Bx + C = 0,\ (A \ne 0)$$ for some constant $$\delta $$, then the value of $$\delta$$ is

A
$$\displaystyle \delta = {1 \over 2}\left( {{B \over A} - {b \over a}} \right)$$

B
$$\displaystyle \delta = {1 \over 2}\left( {{b \over a} - {B \over A}} \right)$$

C
$$\displaystyle {{{b^2} - 4ac} \over {{a^2}}} = {{{B^2} - 4AC} \over {{A^2}}}$$

D
$$\displaystyle {{{b^2} + 4ac} \over {{a^2}}} = {{{B^2} + 4AC} \over {{A^2}}}$$

Solution
Question 6
1 / -0

For $$a \le 0$$, the roots of the equation $${x^2} - 2a\left| {x - a} \right| - 3{a^2} = 0$$ is

A
$$ - a\left( {\sqrt 6 + 1} \right)$$

B
$$a\left( {\sqrt 6 + 1} \right)$$

C
$$a\left( {1 - \sqrt 2 } \right),a\left( { - 1 + \sqrt 6 } \right)$$

D
$$a\left( {1 -\sqrt 2 } \right)$$

Solution

$$x^2-2a|x-a|-3a^2=0$$
Case $$I$$
$$x\ge a$$
$$\implies x^2-2a(x-a)-3a^2=0$$
$$\implies x^2-2ax+a^2=2a^2$$
$$\implies x-a=\pm \sqrt{2}a$$
$$\implies x=a\pm \sqrt{2}a$$
But, $$a<0$$
$$\therefore x=a(1-\sqrt{2})$$
Case $$II$$
as, $$x<0$$, $$a<0$$
But we want only positive roots, we have nothing to do with this case.
Thus, $$x=a(1-\sqrt{2})$$
Question 7
1 / -0

The value of $$'a'$$ for which the quadratic equation $$2{x^2} - x\left( {{a^2} + 8a - 1} \right) + {a^2} - 4a = 0$$ has roots opposite signs, lie in the interval

A
$$1 < a < 5$$

B
$$0 < a < 4$$

C
$$ - 1 < a < 2$$

D
$$2 < a < 6$$

Solution

R.E.F image
Given equation is : $$ 2x^{2}-x(a^{2}+8a-1)+a^{2}-4a = 0 $$
Comparing it with the equation : $$ ax^{2}+bx+c = 0 $$
we have $$ a = 2 $$
$$ b = -(a^{2}+8a-1) $$
$$ c = a^{2}-4a $$
For the roots to have opposite signs.
product of roots $$ < 0 $$
$$ \Rightarrow \dfrac{c}{a}< 0 $$
$$ \Rightarrow \dfrac{a^{2}-4a}{2}< 0 $$
$$ \Rightarrow a^{2}-4a< 0 $$
$$ \Rightarrow a(a-4)< 0 $$
$$ \Rightarrow (a-0)(a-4)< 0 $$
$$ \therefore aE(0,4) $$
Question 8
1 / -0

Find the roots of the equation $$5x^{2} -6x-2 = 0$$ by the method of completing the square.

A
$$\dfrac{3\pm\sqrt{76}}{2}$$

B
$$\dfrac{3\pm\sqrt{19}}{4}$$

C
$$\dfrac{3\pm\sqrt{76}}{10}$$

D
$$\dfrac{3\pm\sqrt{19}}{5}$$

Solution

$$5x^2-6x-2=0\Rightarrow x^2-\dfrac65x-\dfrac25=0\Rightarrow x^2-\dfrac65x=\dfrac25\Rightarrow x^2-2\times \dfrac35x+(\dfrac{3}{5})^2$$
$$\Rightarrow (x-\dfrac35)^2=\dfrac{10}{25}+\dfrac{9}{25}\\\Rightarrow x-\dfrac35=\pm\dfrac{\sqrt{19}}{5}\Rightarrow x=\dfrac{\pm\sqrt{19}+3}{5}$$
Question 9
1 / -0

The roots of the given equation $$\left( {p - q} \right){x^2} + \left( {q - r} \right)x + \left( {r - p} \right) = 0$$ are

A
$$\dfrac{{p - q}}{{r - p}},1$$

B
$$\dfrac{{q - r}}{{p - q}},1$$

C
$$\dfrac{{r - p}}{{p - q}},1$$

D
$$1,\dfrac{{q - r}}{{p - q}}$$

Solution

$$(p-q)x^2+(q-r)x+(r-p)=0$$
$$\implies px^2-qx^2+qx-rx+r-p=0$$
$$\implies p(x^2-1)+q(x-x^2)+r(1-x)=0$$
$$\implies p(x+1)-qx-r=0$$
$$\implies px+p-qx-r=0$$
$$\implies x(p-q)=r-p$$
$$\implies x=\cfrac{r-p}{p-q}$$
$$\therefore x=\cfrac{r-p}{p-q},1$$
Question 10
1 / -0

Two water taps together can fill a tank in $$3\dfrac {1}{13}$$ hours. The tap of longer diameter takes $$3$$ hours less then smaller one to fill the bank separately. Find the time(hrs) in which tap of smaller diameter can separately fill the tank.

A
$$5$$

B
$$3$$

C
$$8$$

D
None of these

Solution

Let time taken by tap of longer diameter be $$x$$ hours

Then time taken by tap of smaller diameter will be $$x+3$$ hours

Now, According to question

$$\dfrac {1}{x} + \dfrac {1}{{x + 3}} = \dfrac {{13}}{{40}}$$

$$ \Rightarrow \dfrac {{2x + 3}}{{x\left( {x + 3} \right)}} = \dfrac {{13}}{{40}}$$

$$ \Rightarrow 80x + 120 = 13{x^2} + 39x$$

$$ \Rightarrow 13{x^2} - 41x - 120 = 0$$

$$ \Rightarrow 13{x^2} + 24x - 65x - 120 = 0$$

$$ \Rightarrow x\left( {13x + 24} \right) - 5\left( {13x + 24} \right) = 0$$

$$ \Rightarrow \left( {13x + 24} \right)\left( {x - 5} \right) = 0$$

$$ \Rightarrow x = \dfrac {{ - 24}}{{13}}$$ or $$x=5$$

Since time cannot be negative
Therefore $$x=5$$

Hence, Time taken by tap of larger diameter is$$5$$ hours and time taken by tap of smaller diameter is $$8$$ hours

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Quadratic Equations Test - 37

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Quadratic Equations Test - 37

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Question 1

$$exist, \sqrt2$$, $$1$$

$$exist, -\sqrt2$$, $$1$$

$$exist, -\sqrt2$$, $$-1$$

$$\text{does not exist}$$

Solution

Question 2

$$4{a^2} + 2a - 2$$

$$-4{a^2} + 2a + 2$$

$$2\left( {2{a^2} + a - 1} \right)$$

$$4{a^2} + 2a + 2$$

Solution

Question 3

An expression with a single variable and degree $$2$$.

An equation with a single variable and degree $$2$$.

An equation with degree $$2$$ only.

An equation with single variable only.

Solution

Question 4

Factorisation Method

Completing Square Method

Hit and Trial Method

All the above.

Solution

Question 5

$$\displaystyle \delta = {1 \over 2}\left( {{B \over A} - {b \over a}} \right)$$

$$\displaystyle \delta = {1 \over 2}\left( {{b \over a} - {B \over A}} \right)$$

$$\displaystyle {{{b^2} - 4ac} \over {{a^2}}} = {{{B^2} - 4AC} \over {{A^2}}}$$

$$\displaystyle {{{b^2} + 4ac} \over {{a^2}}} = {{{B^2} + 4AC} \over {{A^2}}}$$

Solution

Question 6

$$ - a\left( {\sqrt 6 + 1} \right)$$

$$a\left( {\sqrt 6 + 1} \right)$$

$$a\left( {1 - \sqrt 2 } \right),a\left( { - 1 + \sqrt 6 } \right)$$

$$a\left( {1 -\sqrt 2 } \right)$$

Solution

Question 7

$$1 < a < 5$$

$$0 < a < 4$$

$$ - 1 < a < 2$$

$$2 < a < 6$$

Solution

Question 8

$$\dfrac{3\pm\sqrt{76}}{2}$$

$$\dfrac{3\pm\sqrt{19}}{4}$$

$$\dfrac{3\pm\sqrt{76}}{10}$$

$$\dfrac{3\pm\sqrt{19}}{5}$$

Solution

Question 9

$$\dfrac{{p - q}}{{r - p}},1$$

$$\dfrac{{q - r}}{{p - q}},1$$

$$\dfrac{{r - p}}{{p - q}},1$$

$$1,\dfrac{{q - r}}{{p - q}}$$

Solution

Question 10

$$5$$

$$3$$

$$8$$

None of these

Solution

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