IISER Mathematics - Quadratic Equations
Exam Duration: 45 Mins Total Questions : 30
If 1,a1,a2,a3,....an-1 are the nth roots of unity,then value of (1 - a1) (1 - a2)(1 - a3)...(1 - an-1),is
- (a)
n
- (b)
n2
- (c)
n3
- (d)
0
The equation \(\sqrt { x+1 } \)- \(\sqrt { x-1 } \) = \(\sqrt { 4x-1 } \), has
- (a)
no solution
- (b)
one solution
- (c)
two solutions
- (d)
more than two solutions
The number of the real roots of the equation |x2| + 3|x| + 2 = 0, is
- (a)
4
- (b)
3
- (c)
2
- (d)
0
All the terms in the equation are positive, so their sum cannot be zero for any real value of x. Thus the given equation has no real roots. Hence the correct alternative is (d).
A quadratic equation with rational coefficient can have
- (a)
both roots equal and irrational
- (b)
one root rational and other irrational
- (c)
one root real and other imaginary
- (d)
None of these
Irrational and imaginary roots occur in pairs only.
The quadratic equation ax2+bx+c=0 has more than two roots,if
- (a)
\(a\neq 0,b=0,c\neq 0\)
- (b)
\(a\neq 0,b\neq 0,c\neq 0\)
- (c)
\(a=b=c=0\)
- (d)
None of these
A quadratic equation can have more than two roots, if it is an identity, for which a = b = c = O.
The solution of the inequation 2x2+3x-9\(\underline { < } \)0 is given by
- (a)
-3\(\underline { < } \)x\(\underline { < } \)\(\frac { 3 }{ 2 } \)
- (b)
\(\frac { 3 }{ 2 } \)\(\underline { < } \)x\(\underline { < } \)3
- (c)
-3\(\underline { < } \)x\(\underline { < } \)-\(\frac { 3 }{ 2 } \)
- (d)
None of these
2x2+3x-9≤0
2x2+6x-3x-9≤0
(x+3)(2x-3)≤0
∴ (i) x+3≥0 and 2x-3 ≤0
x≥-3 and x≥ \(\frac{3}{2}\) ⇒ -3≤x≤\(\frac{3}{2}\)
i.e. x lies in the interval (-3,\(\frac{3}{2}\))
(ii) x+3 ≤0 and 2x-3≥0
or x≤3 and x≥ \(\frac{3}{2}\); which is not possible.
If roots of the equation lx2+mx+m=0 are in the ratio a:b then the value of \(\sqrt { \frac { a }{ b } } +\sqrt { \frac { b }{ a } } +\sqrt { \frac { m }{ l } } \)is
- (a)
0
- (b)
1
- (c)
3
- (d)
None of these
Let α,β be the roots of the given equation., then
\(\alpha+\beta=\frac{-m}{l}, \alpha\beta=\frac{m}{l}\)
Also, \(\frac{\alpha}{\beta}=\frac{a}{b}\)
Now, \(\sqrt {\frac {a}{b}}+\sqrt {\frac{b}{a}}+\sqrt {\frac{m}{l}}\)
=\(\sqrt {\frac{\alpha}{\beta}}+\sqrt {\frac{\beta}{\alpha}}+\sqrt {\frac{m}{l}}\)
=\(\frac{\alpha+\beta}{\sqrt{\alpha\beta}}+\sqrt \frac{m}{l}\)=\(\frac { \frac { -m }{ l } }{ \sqrt { \frac { m }{ l } } } =\sqrt { \frac { m }{ l } } \)
=-\(\sqrt {\frac{m}{l}}+\sqrt {\frac{m}{l}}=0\)
If a,b,c,d are positive real numbers such that a+b+c+d=2 then M=(a+b)(c+d) satisfies the relation
- (a)
\(0\underline { < } M\underline { < } 1\)
- (b)
\(1\underline { < } M\underline { < } 2\)
- (c)
\(2\underline { < } M\underline { < } 3\)
- (d)
\(3\underline { < } M\underline { < } 4\)
Let a+b and c+d be the roots of the quadratic equation; then
S=sum of the roots = a+b+c+d=2
p= product of the roots = (a+b)(c+d)=M
The quadratic equation is
x2-Sx+P=0
or x2-2x+M=0 --- (i)
Roots of (i) are real, if
4-4M≥0 (∵ b2-4ac≥0)
or 4≥4M
⇒ 4M≤4 ⇒ M≤1
Also, a,b,c,d are positive; thus M≥0
∴ 0≤M≤1
Hence the correct alternatives is (a).
Let \(\alpha ,\beta \) be the roots of the equation,
(x-a)(x-b)+c=0,(c\(\neq \)0).
The roots of the equation \((x-\alpha )(x-\beta )-c=0\) are
- (a)
a,c
- (b)
b,c
- (c)
a,b
- (d)
a+c,b+c
Given equation is
x2-(a+b)x+ab+c=0
∴ α+β=a+b
αβ=ab+c
or αβ-c=ab --- (i)
The equation x2-(α+β)x+αβ-c=0
or x2-(a+b)x+ab=0
⇒ (x-a)(x-b)=0
Thus, roots are x=a,b
Hence the correct alternative is (c).
The value of 'a' for which one root of the quadratic equation (a2-5a+3)x2+(3a-1)x+2=0 is twice as large as the other, is
- (a)
\(\frac { 2 }{ 3 } \)
- (b)
\(-\frac { 2 }{ 3 } \)
- (c)
\(\frac { 1 }{ 3 } \)
- (d)
\(-\frac { 1 }{ 3 } \)
If the roots of the equation ax2+bx+c=0 are in the ratio p:q then
- (a)
pqa2=(p+q)c2
- (b)
pqb2=(p+q)ac
- (c)
pqb2=(p+q)2ac
- (d)
None of these
Let roots be pα,qα
(p+q)α=-\(\frac{b}{a}\) ---(i)
pqα2=\(\frac{c}{a}\) --- (ii)
Substituting value of a from (i) in (ii), we get
pq.\(\frac{b^{2}} {a^{2}(p+q)^{2}} = \frac{c}{a}\)
or pqb2=(p+q)2ac
If the roots of the given equation \(\left( 2k+1 \right) { x }^{ 2 }-\left( 7k+3 \right) x+k+2=0\) are reciprocal to each other, then the value of k will be
- (a)
0
- (b)
1
- (c)
2
- (d)
3
Let roots be \(\alpha \) and \(1/\alpha \) , then
\(\alpha .\frac { 1 }{ \alpha } =\frac { k+2 }{ 2k+1 } \Rightarrow 1=\frac { k+2 }{ 2k+1 } \Rightarrow k=1\)
The equation of the smallest degree with real coefficients having 1+i as one of the root is.
- (a)
\({ x }^{ 2 }+x+1=0\)
- (b)
\({ x }^{ 2 }+2x+2=0\)
- (c)
\({ x }^{ 2 }+2x+2=0\)
- (d)
\({ x }^{ 2 }+2x-2=0\)
Let \(x=1+i\Rightarrow x-1=i\\ \Rightarrow { (x-1) }^{ 2 }={ i }^{ 2 }\Rightarrow { x }^{ 2 }-2x+1=-1\Rightarrow { x }^{ 2 }-2x+2=0\)
Alternate Method
We know that complex roots always occurs in pairs.
Thus, other root will be (1 - i).
Now, sum of roots = 1 + i + 1 - i = 2
Product of roots = (1 + i)(1 - i)=2
Required equation = x2 - 2x + 2
The value of \(2+\frac { 1 }{ 2+\frac { 1 }{ 2+...\infty } } \) is
- (a)
\(1-\sqrt { 2 } \)
- (b)
\(1+\sqrt { 2 } \)
- (c)
\(1\pm \sqrt { 2 } \)
- (d)
None of these
Let x=2+\(\frac{1}{2+\frac{1}{2+..\infty}} \Rightarrow x=2+\frac{1}{x}\) [on simplification]
⇒ x2-2x-1=0 ⇒ x=1士\(\sqrt{2}\)
But the value of the given expression cannot be negative or less than 2, therefore 1+\(\sqrt{2}\) is required answer.
The number of roots of the equation \({ \left| x \right| }^{ 2 }-7\left| x \right| +12=0\) is
- (a)
1
- (b)
2
- (c)
3
- (d)
None of these
Given, equation reduces to (|x|-4(|x|-3)=0
⇒ |x|=4 and |x|=3
⇒ |x|=± 4 and x=±3
The number of solutions for the equation \({ x }^{ 2 }-5\left| x \right| +6=0\) is
- (a)
4
- (b)
3
- (c)
2
- (d)
1
Given equation x2-5|x|+6=0.
When x≥0, x2-5x+6=0
and when x<0, x2+5x+6=0
⇒ x2-3x-2x+6=0; x≥0
and x2+3x+2x+6=0; x<0
⇒ (x-3)(x-2)=0, x≥0 and (x+3).(x+2)=0, x<0
∴ x=3,x=2 and x=-3,x=-2
There are four solution of this equation.
The number of real solutions of the equation \(\left| { x }^{ 2 }+4x+3 \right| +2x+5=0\) are
- (a)
1
- (b)
2
- (c)
3
- (d)
4
Here, two cases arise,
Case I x2 +4x+3>0
This gives, x2+4x+3+2x+5=0
⇒ x2+6x+8=0
⇒ (x+2)(x+4)=0 ⇒ x=-2,-4
x=-2 is not atisfying the condition x2+4x+3>0.
So, x=-4 is the only solution of the given equation.
Case II x2+4x+3<0
This gives, -(x2+4x+3)+2x+5=0
⇒ -x2-2x+2=0 ⇒ x2+2x-2=0
⇒ (x+1-\(\sqrt{3}\))(x+1+\(\sqrt{3}\))=0
∴ x=-1+\(\sqrt{3}\), -1-\(\sqrt{3}\)
Hence x=-(1+\(\sqrt{3}\)) satisfy the given condition x2+4x-3<0, while x=-1+\(\sqrt{3}\) is not satisfying the condition. Thus, number of real solution are two.
If \(\alpha ,\beta \) are the roots of \({ x }^{ 2 }+px+1=0\) and \(\gamma ,\delta \) are the roots of \({ x }^{ 2 }+qx+1=0\) then \({ q }^{ 2 }-{ p }^{ 2 }\) is equal to
- (a)
\(\left( \alpha -\gamma \right) \left( \beta -\gamma \right) \left( \alpha +\delta \right) \left( \beta +\delta \right) \)
- (b)
\(\left( \alpha +\gamma \right) \left( \beta +\gamma \right) \left( \alpha -\delta \right) \left( \beta -\delta \right) \)
- (c)
\(\left( \alpha +\gamma \right) \left( \beta +\gamma \right) \left( \alpha +\delta \right) \left( \beta +\delta \right) \)
- (d)
None of the above
As given, α+β= -p, αβ=1, \(\gamma\)+\(\delta\)=-q and \(\gamma\delta=1\)
Now, \((\alpha-\gamma)(\beta-\gamma)(\alpha+\delta)(\beta+\delta)\)
= {\({\alpha\beta-\gamma(\alpha+\beta)+{\gamma^{2}}}{\alpha\beta+\delta(\alpha+\beta)+{\delta^{2}}}\)}
=\((1+p\gamma+\gamma^{2})(1-p\delta+{\delta^{2}})={(p\gamma-q\gamma)}-{(-p\delta-q\delta)}\)
[Since, \(\gamma and \delta\) are the roots of x2+qx+1=0, therefore \(\gamma^{2}\)+1=-q\(\gamma\) and \(\delta^{2}+1=-q\delta\)
=\(\gamma\delta(q-p)(p+q)=q^{2}-p^{2}\)
If \({ \alpha }\)and \(\beta \) are the roots of the equation \({ x }^{ 2 }-4x+1=0\), then the value of \({ \alpha }^{ 3 }+{ \beta }^{ 3 }\) is
- (a)
76
- (b)
52
- (c)
- 52
- (d)
- 76
\({ \alpha }^{ 3 }+{ \beta }^{ 3 }={ (\alpha +\beta ) }^{ 3 }-3\alpha \beta (\alpha +\beta )={ (4) }^{ 3 }-3\times 1(4)=52[\because \alpha +\beta =4,\alpha \beta =1]\)
If x2 + 5 = 2x - 4 cos (a + b), where \(a,b\epsilon (0,5),\) is satisfied for atleast one real x, then the maximum value of a + b in \([0,2\pi ]\)
- (a)
\(3\pi \)
- (b)
\(2\pi \)
- (c)
\(\pi \)
- (d)
None of the above
x2+5=2x-4 cos(a+b) ⇒ x2-2x+1+4=-4 cos(a+b)
⇒ (x-1)2+4[1+cos(a+b)]=0
⇒ x=1 and 1+cos(a+b)=0
⇒ cos(a+)=-1
⇒ a+b=π,3π,..
If \(a+b+c>\frac { 9c }{ 4 } \) and equation \({ ax }^{ 2 }+2bx-5c=0\) has non - real complex roots, then
- (a)
a > 0, c > 0
- (b)
a > 0, c < 0
- (c)
a < 0, c < 0
- (d)
a < 0, c > 0
Given, 4a+4b-5c>0
Let f(x)=ax2+2bx-5c, then f(2)=4a+4b-5c>0
Since , equation f(x)=0 has imaginary roots, therefore f(x) will have same sign as that of a for all x∈R.
Since, f(2)>0 ⇒ a>0 and c<0
If P(x) = ax2 + bx + c and Q(x) = - ax2 + dx + c, where \(ac\neq 0\) then P(x).Q(x) = 0 has
- (a)
exactly one real root
- (b)
atleast two real roots
- (c)
exactly three real roots
- (d)
all four real roots
D1 : b2 - 4ac and D2 : d2 + 4ac. Hence, atleast one of either D1 or D2 is grater than 0.
Sachin and Rahul attempted to solve a quadratic equation. Sachin made a mistake in writing down the constant term and ended up in roots (4, 3). Rahul made a mistake in writing down coefficient of x to get roots (3, 2). The correct roots of equation are
- (a)
-4, -3
- (b)
6, 1
- (c)
4, 3
- (d)
-6, -1
Let the quadratic equation be ax2 + bx + c = 0
Sachin made a mistake in writing down constant terms.
Therefore, Sum of roots is correct. i.e., \(\alpha +\beta =7\)
Rahul made mistake in writing down coefficient of x
Therefore, Product of roots is correct. i.e.,
\(\Rightarrow Correct\quad quadratic\quad equation\quad is\quad { x }^{ 2 }-(\alpha +\beta )x+\alpha \beta =0.\\ \Rightarrow { x }^{ 2 }-7x+6=0\quad having\quad roots\quad 1\quad and\quad 6.\)
If the roots of the equation bx2 + cx + a = 0 are imaginary, then for all real values of x, the expression \({ 3b }^{ 2 }{ x }^{ 2 }+6bcx+{ 2c }^{ 2 }\) is
- (a)
greater than 4ab
- (b)
less than 4ab
- (c)
greater than - 4ab
- (d)
less than - 4ab
Given, bx2+cx+a=0 has imaginary roots.
⇒ c2-4ab<0 ⇒ c2<4ab ⇒ c2>-4ab --- (i)
Let, f(x)=3b2x2+6bcx+2c2
Here, 3b2>0
So, the given expression has a minimum value.
∴ Minimum value = \(\frac{-D}{4a}=\frac{4ac-b^{2}}{4a}\)
=\(\frac { 4(3b)^{ 2 }.(2c)^{ 2 }-36b^{ 2 }c^{ 2 } }{ 4(3b)^{ 2 } } =-\frac { 12{ b }^{ 2 }{ c }^{ 2 } }{ 12{ b }^{ 2 } } \)
= -c2>-4ab [from Eq.(i)]
Find the value of P such that the sum of the squares of the roots of the equation x2-(a-2)x-(a+1)=0 is least
- (a)
4
- (b)
2
- (c)
1
- (d)
3
Let∝, β br the roots of the equation
∴ ∝+β=a-2 and ∝β=(a+1)
Now, ∝2+β2=(∝+β)2-2∝β=(a-2)2+2(a+1)=(a-1)2+5
∴ ∝2+β2 will be minimum if (a-1)2=0,i.e., a=1
The equations ax2+bx+a=0(a,b∈R) and x3-2x2+2x-1=0 have 2 roots common, Then a+b must be eqaul to
- (a)
1
- (b)
-1
- (c)
0
- (d)
None of these
We have, x3-2x2+2x-1=0
⇒(x-1) (x2-x+1)=0
⇒ x=1, \(\frac { 1\pm \sqrt { 3 } i }{ 2 } \)
Since coefficints of ax2+bx+a=0 are real,
∴ The common roots must be complex and conjugate.
Now, \(\frac { 1+i\sqrt { 3 } }{ 2 } =-\left( \frac { -1-i\sqrt { 3 } }{ 2 } \right) =-{ \omega }^{ 2 }\)
substituting x=-ω2 in ax2+bx+a=0
⇒aω4-bω2+a=0⇒aω-bω2+a=0
⇒a(1+ω)-bω2=0⇒a(-ω2)-bω2=0
⇒-ω2(a+b)=0⇒a+b=0
If the roots of the equation \(\frac{a}{x-a}+\frac{b}{x-b}=1\) are equal in magnitude and opposite in sign, then
- (a)
a=b
- (b)
a+b=1
- (c)
a-b=1
- (d)
a+b=0
Given equation is \(\frac { a }{ x-a } +\frac { b }{ x-b } =1\)
⇒a(x-b)+b(x-a)=(x-a)(x-b)
⇒x2-x(a+b) +ab=ax-ab+bx-ab
⇒x2-2x(a+b)+3ab=0
So, sum of roots =∝+(-∝)=2(a+b)
or a+b=0.
If 2+i is a root of equation x3-5x2+9x-5=0, then the other roots are to
- (a)
1 and 2-i
- (b)
-1 and 3+i
- (c)
0 and 1
- (d)
-1 and i-2
Since complex roots always occur in pairs
So, 2-i is also a root of given equation.
Given equation is x3-5x2+9x-5=0.
x=1 satisfies this equation.
∴Other roots are (2-i) and 1.
If ∝ and β are the roots of the equation (x-a)(x-b)=5, then the roots of the equation (x-∝)(x-β)+5=0 are
- (a)
a,5
- (b)
b,5
- (c)
a,∝
- (d)
a,b
Since ∝,βare roots of the equation (x-a)(x-b)=5 or x2-(a+b)x+(ab-5)=0
Taking another equation (x-∝)(x-β)+5=0
⇒x2-(∝+β)x+(∝β+5)=0
⇒x2-(a+b)x+ab=0 (Using (i))
∴ Its roots are a,b.
If 3P2=5P+2 and 3q2=5q+2 where P≠q, then the equation whose roots are 3p-2q and 3q-2p is
- (a)
3x2-5x-100=0
- (b)
5x2+3x+100=0
- (c)
3x2-5x+100=0
- (d)
3x2+5x=100=0
3p2=5p+2⇒3p2-5p-2=0
⇒(3p+1)(p-2)=0 ∴p=-1/3, 2.
3q2=5q+2⇒3q2-5q-2=0 ∴q=-1/3,2.
It's given that P≠q
So, if p=2, q should be -1/3 or vice versa.
Now, roots are (3P-2q) and (3q-2p)
∝=3p-2q=3x2-2\(\left( -\frac { 1 }{ 3 } \right) \)=6+\(\frac{2}{3}=\frac{20}{3}\)
and β=3q-2p=3\(\left( -\frac { 1 }{ 3 } \right) \)-2x2=-5
Now ∝+β=\(\frac{20}{3}\)+(-5)=\(\frac{5}{3}\)and ∝β=\(\frac{20}{3}\)x(-5)=-\(\frac{100}{3}\)
Now required equation is x2-(∝+β)x+∝β=0
∴x2-\(\frac{5}{3}x-\frac{100}{3}=0\)