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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Question 1

exist,2exist, \sqrt2exist,2​, 111

exist,−2exist, -\sqrt2exist,−2​, 111

exist,−2exist, -\sqrt2exist,−2​, −1-1−1

does not exist\text{does not exist}does not exist

Solution

Question 2

4a2+2a−24{a^2} + 2a - 24a2+2a−2

−4a2+2a+2-4{a^2} + 2a + 2−4a2+2a+2

2(2a2+a−1)2\left( {2{a^2} + a - 1} \right)2(2a2+a−1)

4a2+2a+24{a^2} + 2a + 24a2+2a+2

Solution

Question 3

An expression with a single variable and degree 222.

An equation with a single variable and degree 222.

An equation with degree 222 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

δ=12(BA−ba)\displaystyle \delta = {1 \over 2}\left( {{B \over A} - {b \over a}} \right)δ=21​(AB​−ab​)

δ=12(ba−BA)\displaystyle \delta = {1 \over 2}\left( {{b \over a} - {B \over A}} \right)δ=21​(ab​−AB​)

b2−4aca2=B2−4ACA2\displaystyle {{{b^2} - 4ac} \over {{a^2}}} = {{{B^2} - 4AC} \over {{A^2}}}a2b2−4ac​=A2B2−4AC​

b2+4aca2=B2+4ACA2\displaystyle {{{b^2} + 4ac} \over {{a^2}}} = {{{B^2} + 4AC} \over {{A^2}}}a2b2+4ac​=A2B2+4AC​

Solution

Question 6

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

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

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

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

Solution

Question 7

1<a<51 < a < 51<a<5

0<a<40 < a < 40<a<4

−1<a<2 - 1 < a < 2−1<a<2

2<a<62 < a < 62<a<6

Solution

Question 8

3±762\dfrac{3\pm\sqrt{76}}{2}23±76​​

3±194\dfrac{3\pm\sqrt{19}}{4}43±19​​

3±7610\dfrac{3\pm\sqrt{76}}{10}103±76​​

3±195\dfrac{3\pm\sqrt{19}}{5}53±19​​

Solution

Question 9

p−qr−p,1\dfrac{{p - q}}{{r - p}},1r−pp−q​,1

q−rp−q,1\dfrac{{q - r}}{{p - q}},1p−qq−r​,1

r−pp−q,1\dfrac{{r - p}}{{p - q}},1p−qr−p​,1

1,q−rp−q1,\dfrac{{q - r}}{{p - q}}1,p−qq−r​

Solution

Question 10

555

333

888

None of these

Solution

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$exist, \sqrt2$ , $1$

$exist, -\sqrt2$ , $1$

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

$\text{does not exist}$

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

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

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

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

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

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

An equation with degree $2$ only.

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

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

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

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

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

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

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

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

$1 < a < 5$

$0 < a < 4$

$- 1 < a < 2$

$2 < a < 6$

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

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

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

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

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

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

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

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

$5$

$3$

$8$