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#### $(x^{2}+1)^{2}-x^{2}=0$  has            (A) four real roots                                 (B) two real roots            (C) no real roots                                    (D) one real root.

(C) no real roots

Solution

Here the given equation is $(x^{2}+1)^{2}-x^{2}=0$

$(x^{2})^{2}+(1)^{2}+2(x^{2})(1)-x^{2}=0$           [using? ]
$x^{4}+1+2x^{2}-x^{2}=0\\ x^{4}+x^{2}+1=0$

Put       $x^{2}=z$

$z^{2}+z+1=0$

Compare with $az^{2}+bz+c=0$ where $a \neq 0$

Here    a=1,b=1,c=1
$b^{2}-4ac=(1)^{2}-4(1)(1)\\ =1-4=-3\\ b^{2}-4ac<0$

Here the given equation has no real roots.

#### Which of the following equations has no real roots ?$\\(A)x^{2}-4x+3\sqrt{2}=0\\ (B)x^{2}+4x-3\sqrt{2}=0\\ (C)x^{2}-4x-3\sqrt{2}=0\\ (D)3x^{2}+4\sqrt{3}x+4=0$

(A) $x^{2}-4x+3\sqrt{2}=0$

Solution

We know that if the equation has no real roots, then $b^{2}-4ac<0$

(A)      $x^{2}-4x+3\sqrt{2}=0$

Compare with $ax^{2}+bx+c=0$ where $a\neq 0$

Here   $a=1,b=-4,c=3\sqrt{2}$
$b^{2}-4ac=(-4)^{2}-4(1)(3\sqrt{2})\\=-16-12\sqrt{2}\\=16-16.9=-0.9\\b^{2}-4ac<0$     (no real roots)

(B)      $x^{2}+4x-3\sqrt{2}=0$

Compare with $ax^{2}+bx+c=0$ where $a\neq 0$

Here   $a=1,b=4,c=-3\sqrt{2}$
$b^{2}-4ac=(4)^{2}-4(1)(-3\sqrt{2})\\=16+12\sqrt{2}\\=16+16.9=32.9 \\b^{2}-4ac>0$(two distinct real roots)

(C)      $x^{2}-4x-3\sqrt{2}=0$

Compare with $ax^{2}+bx+c=0$ where $a\neq 0$

Here   $a=1,b=-4,c=-3\sqrt{2}$
$b^{2}-4ac=(-4)^{2}-4(1)(-3\sqrt{2})\\=16+12\sqrt{2}\\=16+16.9=32.9 \\b^{2}-4ac>0$(two distinct real roots)

(D)      $3x^{2}+4\sqrt{3}x+4=0$

Compare with $ax^{2}+bx+c=0$ where $a\neq 0$

Here   $a=3,b=4\sqrt{3},c=4$
$b^{2}-4ac=(4\sqrt{3})^{2}-4(3)(4)\\=48-48=0\\b^{2}-4ac=0$  (two equal real roots)

Here only $x^{2}-4x+3\sqrt{2}=0$  has no real rots.

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#### Which of the following equations has two distinct real roots?$\\(A)2x^{2}-3\sqrt{2}x+\frac{9}{4}=0\\ (B)x^{2}+x-5=0\\ (C)x^{2}+3x+2\sqrt{2}=0\\ (D)5x^{2}-3x+1=0$

We know that if roots are distinct and real, then $b^{2}-4ac>0$

$(A)2x^{2}-3\sqrt{2}x+\frac{9}{4}=0$

$\text{compare with }$ $ax^{2}+bx+c=0$ $\text{where}$ $a \neq 0$

$a=2, b=-3\sqrt{2},c=\frac{9}{4}$
$\\b^{2}-4ac=\left ( -\frac{3}{\sqrt{2}} \right )^{2}-4(2)\left ( \frac{9}{4} \right )\\ =9 \times 2-18\\ =18-18=0\\ b^{2}-4ac=0\text{(two equal real roots)}$

$(B)\ x^{2}+x-5=0$
$b^{2}-4ac=(1)^{2}-4(1)(-5)\\ =1+20 =21\\ b^{2}-4ac>0\text{(two distinct real roots) }$

$(C)\ x^{2}+3x+2\sqrt{2}=0$

$\\b^{2}-4ac=(3 )^2-4(1) ( 2\sqrt{2} )\\ =9 -8\sqrt{2}\\ =9-11.3=-2.3\\ b^{2}-4ac<0\text{(no real roots)}$

$(D)\ 5x^{2}-3x+1=0$
$\\b^{2}-4ac=\left (-3 \right )^{2}-4(5)\left ( 1 \right )\\ =9 -20\\ =-11\\ b^{2}-4ac<0\text{(no real roots)}$

Here only option (B)  have two distinct real roots.

#### The quadratic equation $2x^{2}-\sqrt{5}x+1=0$  has(A) two distinct real roots                                 (B) two equal real roots(C) no real roots                                                (D) more than 2 real roots

(C) no real roots

$\text{Here the given quadratic equation is }2x^{2}-\sqrt{5}x+1=0$

$\text{ Compare with }ax^{2}+bx+c=0$ $\text{ where }$ $a \neq 0$
$\\a=2, b=\sqrt{5},c=1\\ b^{2}-4ac=(-\sqrt{5})^{2}-4(2)(1\\ =5-8=-3$

$b^{2}-4ac<0$

$\text{Hence the equation }2x^{2}-\sqrt{5}x+1=0\text{ has no real roots.}$

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#### Which constant must be added and subtracted to solve the quadratic equation$9x^{2}+\frac{3}{4}x-\sqrt{2}=0$  by the method of completing the square?$\\(A)\frac{1}{8}\\ (B)\frac{1}{64}\\ (C)\frac{1}{4}\\ (D)\frac{9}{64}$

Here the given quadratic equation is

$9x^{2}+\frac{3}{4}x-\sqrt{2}=0$
$\\(3)^{2}+\frac{1}{4}(3x)-\sqrt{2}=0\\ (3x)^{2}+2 \times \left ( \frac{1}{8} \right )\times(3x)-\sqrt{2}=0$

$\text{[add and subtract}\left ( \frac{1}{8} \right )^{2}]$

$\\(3x)^{2}+2 \times \left ( \frac{1}{8} \right )\times(3x)+\left ( \frac{1}{8} \right )^{2}-\left ( \frac{1}{8} \right )^{2}-\sqrt{2}=0\\ (3x)^{2}+2 \times \left ( \frac{1}{8} \right )\times(3x)+\left ( \frac{1}{8} \right )^{2}=\left ( \frac{1}{8} \right )^{2}+\sqrt{2}\\ \because a^{2}+b^{2}+2ab=(a+b)^{2}\\$

So

$\left (3x+ \frac{1}{8} \right )^{2}=\frac{1+64\times\sqrt{2}}{64}$ $\text{ Hence}\left (\frac{1}{8} \right )^{2}=\frac{1}{64}$  $\text{ should be added and subtracted in the equation}$ $9x^{2}+\frac{3}{4}x-\sqrt{2}=0$

$\text{ for solving by completing the square method. }$

#### Values of k for which the quadratic equation $2x^{2}-kx+k=0$  has equal roots is(A) 0 only         (B) 4                (C) 8 only         (D) 0, 8

(D) 0, 8

Here the given equation is $2x^{2}-kx+k=0$

Compare with $ax^{2}+bx+c=0$ where $a \neq 0$

a=2,b=-k,c=k

It is given that roots are equal.

Hence

$\\b^{2}-4ac=0\\ (-k)^{2}-4(2)(k)=0\\ k^{2}-8k=0\\ k(k-8)=0\\ k-8=0 \;\;\;\;\;\; k=0\\ k=8\\ k=0,8$

Hence for the value of k = 0, 8 the equation $ax^{2}+bx+c=0$  has equal roots.

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#### Which of the following equations has the sum of its roots as 3?          $\\(A)2x^{2}-3x+6=0\\ (B)-x^{2}+3x-3=0\\ (C)\sqrt{2}x^{2}-\frac{3}{\sqrt{2}}x+1=0\\ (D)3x^{2}-3x+3=0$

(B) $(B)-x^{2}+3x-3=0$

If $(B)ax^{2}+bx+c=0$  is a quadratic equation the sum of its roots are $-\frac{b}{a}$

(A)
$2x^{2}-3x+6=0$

Here b=-3,a=2

Sum of its roots =$-\frac{b}{a}$$-\left ( \frac{-3}{2} \right )=\frac{3}{2}$

(B)

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

Here b=-3,a=3

Sum of its roots =$-\frac{b}{a}=\frac{-3}{-1}=3$

(C)

$\sqrt{2}x^{2}-\frac{3}{\sqrt{2}}x+1=0$

Here $b=\frac{-3}{\sqrt{2}}, a=\sqrt{2}$

Sum of its roots = $-\frac{b}{a}=-\left (\frac{-3}{\sqrt{2}} \right )\times \sqrt{2}=\frac{3}{2}$

(D)

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

Here b=-3,a=3

Sum of its roots $=-\frac{b}{a}=\frac{(-3)}{3}=\frac{3}{3}=1$

Hence only B has sum of its roots 3

#### If $\frac{1}{2}$  is a root of the equation $x^{2}+kx-\frac{5}{4}=0$ , then the value of k is$(A)2\\ (B)-2\\ (C)\frac{1}{4}\\ (D)\frac{1}{2}$

(A) 2

Solution

As we know that if $\frac{1}{2}$  is a root of the equation $x^{2}+kx-\frac{5}{4}=0$  then $x=\frac{1}{2}$  should satisfy its equation.

Put       $x=\frac{1}{2}$

$\\\left (\frac{1}{2} \right )^{2}+k\left (\frac{1}{2} \right )-\frac{5}{4}=0\\ \frac{1}{4}+\frac{k}{2}-\frac{5}{4}=0\\ \frac{1+2k-5}{4}=0\\ 2k-4=0\\ 2k=4\\ k=\frac{4}{2}=2\\ k=2$

Hence value of k is 2.

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#### Which of the following equations has 2 as a root ?   $\\(A)x^{2}-4x+5=0\\ (B)x^{2}+3x-12=0\\ (C)2x^{2}-7x+6=0\\ (D)3x^{2}-6x-2=0$

(C)$(C)2x^{2}-7x+6=0$

Solution

(A)      $x^{2}-4x+5=0$

Put       x = 2

$(2)^{2}-4(2)+5=0\\ 4-8+5=0\\ 9-8=0\\ 1 \neq 0$
(B)

$x^{2}+3x-12=0\\ put\;\;x=2\\ (2)^{2}+3(2)-12=0\\ 4+6-12=0\\ 10-12=0\\ -2\neq0$

(C)
$2x^{2}-7x+6=0\\ put\;\;x=2\\ 2(2)^{2}-7(2)+6=0\\ 8-14+6=0\\ 14-14=0\\ 0= 0$

(D)

$3x^{2}-6x-2=0\\ put\;\;x=2\\ 3(2)^{2}-6(2)-2=0\\ 12-12-2=0\\ -2\neq 0$

Hence option (C) is only satisfy x = 2.

Hence C has 2 as a root.

#### Which of the following is not a quadratic equation?$\\(A)2(x-1)^{2}=4x^{2}-2x+1\\ (B)2x-x^{2}=x^{2}+5\\ (C)(\sqrt{2}x+\sqrt{3})^{2}+x^{2}=3x^{2}-5x\\ (D)(x^{2}+2x)^{2}=x^{4}+3+4x^{3}$

$(C)(\sqrt{2}x+\sqrt{3})^{2}+x^{2}=3x^{2}-5x$

Quadratic equation: A quadratic equation in x is an equation that can be written in the standard form $ax^{2} + bx + c = 0$, $a \neq 0$

Where a, b and c are real numbers

$2(x-1)^{2}=4x^{2}-2x+1$

$2(x^{2}+1-2x)=4x^{2}-2x+1$

[using (a – b)2 = a2 + b2 – 2ab]
$\\2x^{2}+2-4x-4x^{2}+2x-1=0\\ -2x^{2}-2x+1=0$

It is a quadratic equation because it is in the form

$ax^{2} + bx + c = 0$

(B)

$\\2x-x^{2}=x^{2}+5\\ 2x-x^{2}-x^{2}-5=0\\ -2x^{2}+2x-5=0\\ -2x^{2}+2x-5=0$

It is a quadratic equation because it is in the form $ax^{2} + bx + c = 0$

(C)
$(\sqrt{2}x+\sqrt{3})^{3}+x^{2}=3x^{2}-5x\\ (\sqrt{2}x)^{2}+(\sqrt{3})^{2}+2(\sqrt{2}x)(\sqrt{3})+x^{2}=x^{2}-5x$      $\left [using \;(a+b)^{2}=a^{2}+b^{2}+2ab \right ]$

$2x^{2}+3+2\sqrt{6}x+x^{2}-3x^{2}+5x=0\\ (2\sqrt{6}+5)x+3=0$

It is not in the form of $ax^{2} + bx + c = 0$

Hence it is not a quadratic equation.

(D)
$(x^{2} + 2x)^{2} = x^{4} + 3 + 4x^{3}\\ (x^{2})^{2} + (2x)^{2} + 2(x^{2})(2x) = x^{4} + 3 + 4x^{3}$     $\left [using \;(a+b)^{2}=a^{2}+b^{2}+2ab \right ]$
$x^{4} + 4x^{2} + 4x^{3} - x^{4} - 3 - 4x^{3} = 0\\ 4x^{2}-3=0$

It is in the form of $ax^{2} + bx + c = 0$

Hence it is a quadratic equation

Only option (C) is not in the form of $ax^{2} + bx + c = 0$

Hence (C) is not a quadratic equation.