Quadratic Equation Questions for NMAT PDF:
Download Quadratic Equation Questions for NMAT PDF. Top 10 very important Quadratic Equation Questions for NMAT based on asked questions in previous exam papers.
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Question 1:Â The number of integers that satisfy the equality $(x^{2}-5x+7)^{x+1}=1$ is
a)Â 3
b)Â 2
c)Â 4
d)Â 5
Question 2:Â The number of distinct real roots of the equation $(x+\frac{1}{x})^{2}-3(x+\frac{1}{x})+2=0$ equals
Question 3:Â How many disticnt positive integer-valued solutions exist to the equation $(X^{2}-7x+11)^{(X^{2}-13x+42)}=1$ ?
a)Â 8
b)Â 4
c)Â 2
d)Â 6
Question 4:Â If $x^2 + x + 1 = 0$, then $x^{2018} + x^{2019}$ equals which of the following:
a)Â $x+1$
b)Â $x$
c)Â $-x$
d)Â $x-1$
Question 5: If $U^{2}+(U-2V-1)^{2}$= −$4V(U+V)$ , then what is the value of $U+3V$ ?
a)Â $0$
b)Â $\dfrac{1}{2}$
c)Â $\dfrac{-1}{4}$
d)Â $\dfrac{1}{4}$
Question 6:Â If $x+1=x^{2}$ and $x>0$, then $2x^{4}$Â is
a)Â $6+4\sqrt{5}$
b)Â $3+3\sqrt{5}$
c)Â $5+3\sqrt{5}$
d)Â $7+3\sqrt{5}$
Question 7:Â 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)Â 25, 43
b)Â 52, 43
c)Â 52, 67
d)Â None of the above
Question 8:Â If $xy + yz + zx = 0$, then $(x + y + z)^2$ equals
a)Â $(x + y)^2 + xz$
b)Â $(x + z)^2 + xy$
c)Â $x^2 + y^2 + z^2$
d)Â $2(xy + yz + xz)$
Question 9:Â If the equation $x^3 – ax^2 + bx – a = 0$ has three real roots, then it must be the case that,
a)Â b=1
b)Â b $\neq$ 1
c)Â a=1
d)Â a $\neq$ 1
Question 10:Â 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)Â $\frac{-1}{\sqrt 3}$
b)Â $-1$
c)Â $0$
d)Â $1$
e)Â $\frac{1}{\sqrt 3}$
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Answers & Solutions:
1) Answer (A)
$\left(x^2-5x+7\right)^{x+1}=1$
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.
2)Â Answer:Â 1
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
3) Answer (D)
$(X^{2}-7x+11)^{(X^{2}-13x+42)}=1$
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.
4) Answer (C)
We know that,
$x^{3} – 1 = (x – 1)(x^{2} + x + 1)$
Since, $x^2 + x + 1 = 0$
$\therefore $Â $x^{3} – 1$ = 0
=>Â $x^{3}$ = 1
Now, $x^{2018} + x^{2019}$
=Â $(x^{3})^{672} * x^{2}$ +Â $(x^{3})^{673}$
=Â $1^{672} * x^{2}$ +Â $1^{673}$
=Â $x^{2}$ + 1
= -x
Hence, option C.
5) Answer (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.
6) Answer (D)
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, $2x^4 = 6x+4 = 3+3\sqrt{5}+4 = 3\sqrt{5}+7$
7) Answer (C)
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.
8) Answer (C)
$(x+y+z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + xz)$
as $xy+yz+xz = 0$
so equation will be resolved to $x^2 + y^2 + z^2$
9) Answer (B)
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
10) Answer (B)
b = sum of the roots taken 2 at a time.
Let the roots be n-1, n and n+1.
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.
We hope this Quadratic Equation Questions for NMAT pdf for NMAT exam will be highly useful for your Preparation.