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# Quadratic Equation Questions for NMAT PDF:

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Question 1: The number of integers that satisfy the equality $(x^{2}-5x+7)^{x+1}=1$ is

a) 3

b) 2

c) 4

d) 5

Question 2: The number of distinct real roots of the equation $(x+\frac{1}{x})^{2}-3(x+\frac{1}{x})+2=0$ equals

Question 3: How many disticnt positive integer-valued solutions exist to the equation $(X^{2}-7x+11)^{(X^{2}-13x+42)}=1$ ?

a) 8

b) 4

c) 2

d) 6

Question 4: If $x^2 + x + 1 = 0$, then $x^{2018} + x^{2019}$ equals which of the following:

a) $x+1$

b) $x$

c) $-x$

d) $x-1$

Question 5: If $U^{2}+(U-2V-1)^{2}$= −$4V(U+V)$ , then what is the value of $U+3V$ ?

a) $0$

b) $\dfrac{1}{2}$

c) $\dfrac{-1}{4}$

d) $\dfrac{1}{4}$

Question 6: If $x+1=x^{2}$ and $x>0$, then $2x^{4}$  is

a) $6+4\sqrt{5}$

b) $3+3\sqrt{5}$

c) $5+3\sqrt{5}$

d) $7+3\sqrt{5}$

Question 7: If $x^{2}$+3x-10is a factor of $3x^{4}+2x^{3}-ax^{2}+bx-a+b-4$ then the closest approximate values of a and b are

a) 25, 43

b) 52, 43

c) 52, 67

d) None of the above

Question 8: If $xy + yz + zx = 0$, then $(x + y + z)^2$ equals

a) $(x + y)^2 + xz$

b) $(x + z)^2 + xy$

c) $x^2 + y^2 + z^2$

d) $2(xy + yz + xz)$

Question 9: If the equation $x^3 – ax^2 + bx – a = 0$ has three real roots, then it must be the case that,

a) b=1

b) b $\neq$ 1

c) a=1

d) a $\neq$ 1

Question 10: If the roots of the equation $x^3 – ax^2 + bx – c = 0$ are three consecutive integers, then what is the smallest possible value of b?[CAT 2008]

a) $\frac{-1}{\sqrt 3}$

b) $-1$

c) $0$

d) $1$

e) $\frac{1}{\sqrt 3}$

$\left(x^2-5x+7\right)^{x+1}=1$

There can be a solution when $\left(x^2-5x+7\right)=1$ or $x^2-5x\ +6=0$

or x=3 and x=2

There can also be a solution when x+1 = 0 or x=-1

Hence three possible solutions exist.

Let $a=x+\frac{1}{x}$
So, the given equation is $a^2-3a+2=0$
So, $a$ can be either 2 or 1.

If $a=1$, $x+\frac{1}{x}=1$ and it has no real roots.
If $a=2$, $x+\frac{1}{x}=2$ and it has exactly one real root which is $x=1$

So, the total number of distinct real roots of the given equation is 1

$(X^{2}-7x+11)^{(X^{2}-13x+42)}=1$

if $(X^{2}-13x+42)$=0 or $(X^{2}-7x+11)$=1 or $(X^{2}-7x+11)$=-1 and $(X^{2}-13x+42)$ is even number

For X=6,7 the value $(X^{2}-13x+42)$=0

$(X^{2}-7x+11)$=1 for X=5,2.

$(X^{2}-7x+11)$=-1 for X=3,4 and for X=3 or 4, $(X^{2}-13x+42)$ is even number.

.’. {2,3,4,5,6,7} is the solution set of X.

.’. X can take six values.

We know that,

$x^{3} – 1 = (x – 1)(x^{2} + x + 1)$

Since, $x^2 + x + 1 = 0$

$\therefore$ $x^{3} – 1$ = 0

=> $x^{3}$ = 1

Now, $x^{2018} + x^{2019}$

= $(x^{3})^{672} * x^{2}$ + $(x^{3})^{673}$

= $1^{672} * x^{2}$ + $1^{673}$

= $x^{2}$ + 1

= -x

Hence, option C.

Given that $U^{2}+(U-2V-1)^{2}$= −$4V(U+V)$

$\Rightarrow$ $U^{2}+(U-2V-1)(U-2V-1)$= −$4V(U+V)$

$\Rightarrow$ $U^{2}+(U^2-2UV-U-2UV+4V^2+2V-U+2V+1)$ = −$4V(U+V)$

$\Rightarrow$ $U^{2}+(U^2-4UV-2U+4V^2+4V+1)$ = −$4V(U+V)$

$\Rightarrow$ $2U^2-4UV-2U+4V^2+4V+1=−4UV-4V^2$

$\Rightarrow$ $2U^2-2U+8V^2+4V+1=0$

$\Rightarrow$ $2[U^2-U+\dfrac{1}{4}]+8[V^2+\dfrac{V}{2}+\dfrac{1}{16}]=0$

$\Rightarrow$ $2(U-\dfrac{1}{2})^2+8(V+\dfrac{1}{4})^2=0$

Sum of two square terms is zero i.e. individual square term is equal to zero.

$U-\dfrac{1}{2}$ = 0 and $V+\dfrac{1}{4}$ = 0

U = $\dfrac{1}{2}$ and V = $-\dfrac{1}{4}$

Therefore, $U+3V$ = $\dfrac{1}{2}$+$\dfrac{-1*3}{4}$ = $\dfrac{-1}{4}$. Hence, option C is the correct answer.

We know that $x^2 – x – 1=0$
Therefore $x^4 = (x+1)^2 = x^2+2x+1 = x+1 + 2x+1 = 3x+2$
Therefore, $2x^4 = 6x+4$

We know that $x>0$ therefore, we can calculate the value of $x$ to be $\frac{1+\sqrt{5}}{2}$
Hence, $2x^4 = 6x+4 = 3+3\sqrt{5}+4 = 3\sqrt{5}+7$

If $x^{2}$+3x-10is a factor of $3x^{4}+2x^{3}-ax^{2}+bx-a+b-4$
Then x = -5 and x = 2 will give $3x^{4}+2x^{3}-ax^{2}+bx-a+b-4$ = 0
Substituting x = -5 we get,
$3(-5)^{4}+2(-5)^{3}-a(-5)^{2}+b(-5)-a+b-4 = 0$
Solving we get,
$26a+4b = 1621$…….(i)
Substituting x = 2 we get,
$3(2)^{4}+2(2)^{3}-a(2)^{2}+b(2)-a+b-4 =0$
=> $5a-3b = 60$……..(ii)
Solving i and ii we get
a and b $\approx 52, 67$
Hence, option C is the correct answer.

$(x+y+z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + xz)$
as $xy+yz+xz = 0$
so equation will be resolved to $x^2 + y^2 + z^2$

It can be clearly seen that if b=1 then $x^2(x – a) + (x – a) = 0$ an the equation gives only 1 real value of x
Therefore, $b = (n-1)n + n(n+1) + (n+1)(n-1) = n^2 – n + n^2 + n + n^2 – 1$
$b = 3n^2 – 1$. The smallest value is -1.