Number of values `p' for which the equation (p^{2} – 3p + 2) x^{2} – (p^{2} – 5p + 4)x + p – p^{2} = 0 possess more than two roots, is
The roots of the equation (b – c) x^{2} + (c – a) x + (a – b) = 0 are
Clearly, x = 1 is a root of the given equation.
Let, alpha, beta are the roots ; alpha = 1
We know , product of the roots :
Alpha * Beta = ab / bc
1* ( beta) = ab / b c
Beta = ab/ b c
Therefore, roots are : ( ab/bc, 1)
If a, b are the roots of quadratic equation x^{2} + px + q = 0 and g, d are the roots of x^{2} + px – r = 0, then (a – g) . (a  d) is equal to
x^{2}+px+q=0 have roots a,b
a=b=p (1)
ab = q (2)
a2+pa+q=0 (3)
x^{2}+px−r=0 have roots g,d
g+d=p (4)
gd = r (5)
(ag)(ad) = a^{2}−(g−d)a+gd
= a^{2}+pa−r = qr =(q+r) from (3)
Therefore (q+r) id the answer.
Two real numbers a & b are such that a + b = 3 & a  b = 4, then a & b are the roots of the quadratic equation
If a, b, c are integers and b^{2} = 4(ac + 5d^{2}), d ∈ N, then roots of the equation ax^{2} + bx + c = 0 are
b^{2} = 4(ac + 5d^{2})
D = b^{2} − 4ac
= 4(ac + 5d^{2}) − 4ac
= 4ac + 20d^{2} − 4ac
= 20d^{2}
Now, x = (−b ± √D)/2a
=(− b ± √20d^{2})/2a
=(− b ± 2√5d)/2a
Therefore, roots of the given quadratic equation are irrational.
Let a, b and c are real numbers such that 4a + 2b + c = 0 and ab > 0. Then the equation ax^{2} + bx + c = 0 has
Put x=2 in the LHS of the quadratic equation
We get, 4a + 2b + c = 0
Hence x=2 is a root of the equation. (Real roots)
Which of the following graph represents expression f(x) = ax^{2} + bx + c (a < 0) when a > 0, b < 0 & c < 0 ?
The expression y = ax^{2} + bx + c has always the same sign as of `a' if
For the quadratic curve to always have the same sign, it must never cut the Xaxis, i i.e. it must not have any real roots.
So, b^{2} – 4ac < 0
4ac > b^{2}
If a, b ∈ R, a < 0 and the quadratic equation ax^{2} – bx + 1 = 0 has imaginary roots then a + b + 1 is
f(x) = ax^{2}  bx + c
f(0) = 1 (positive)
That is, f(x) = ax^{2}  bx + 1 = 0
Therefore f(x) has imaginary roots
f(1) = a+b+c > 1 (Hence positive)
If a and b are the non–zero distinct roots of x^{2} + ax + b = 0, then the least value of x^{2} + ax + b is
ANSWER : c
Solution : a and b of x^2 + ax + b = 0
Sum of roots : a + b = a, 2a + b = 0
Product of roots : ab = b, b(a1)
Minimum value of x^2 + ax + b at x =a/2
We get a = 9/4
If y = –2x^{2} – 6x + 9, then
y = 2x^{2}  6x + 9
dy/dx = 4x  6
Put dy/dx = 0
4x  6 = 0
x = 3/2 = 1.5
d^{2}y/dx^{2} = 4 < 0 (Maximum point)
Max value = 2(1.5)^{2}  6(1.5) + 9
= 13.5
Max value of x = 1.5
The number of the integer solutions of x^{2} + 9 < (x + 3)^{2} < 8x + 25 is
If both roots of the quadratic equation (2 – x) (x + 1) = p are distinct & positive, then p must lie in the interval
The equatiobn π^{x} = –2x^{2} + 6x – 9 has
The RHS of the expression has a<0 which means the graph will lie below the xaxis and π^{x} lies above the xaxis.Therefore,no solution.
Let a > 0, b > 0 and c > 0. Then both the roots of the equation ax^{2} + bx + c = 0
The set of all solutions of the inequality < 1/4 contains the set
(1/2)^{(x2  2x)} < (1/4)
(1/2)^{(x2  2x)} < (1/2)^{2}
x^{2 }− 2x > 2......(as after multiplicative inverse sign of inequality changes)
x^{2} − 2x − 2 > 0
x^{2}  2x + 1  3 >
(x1)^{2}  3 > 0
(x1)^{2} > 3
So for the above to hold good both the expression must be positive or both must be negative. After finding the solution the range of the solution will be
either x > 3
(3,¥)
Consider y =, where x is real, then the range of expression y^{2} + y – 2 is
The values of k, for which the equation x^{2} + 2(k – 1) x + k + 5 = 0 possess atleast one positive root, are
If <= 4, then least and the highest values of 4x^{2} are
If coefficients of the equation ax^{2} + bx + c = 0, a < 0 are real and roots of the equation are non–real complex and a + c + b < 0, then
Puting x=1
ab+c
but
a+c<b or a+cb<0
hence f(x)<0 for all real values of x
therefore
putting x=2
we get f(X)=4a+c2b<0
or 4a+c<2b
If the roots of the equation x^{2} + 2ax + b = 0 are real and distinct and they differ by at most 2m, then b lies in the interval
If the roots of the quadratic equation x^{2} + px + q = 0 are tan 30º and tan 15º respectively, then the value of 2 + q – p is
If (1 – p) is root of quadratic equation x^{2} + px + (1 – p) = 0, then its roots are
As (1 – p) is root of the equation: x^{2} + px + (1 – p) = 0
(1 – p)^{2} + p(1 – p) + (1 – p) = 0
(1 – p)[1 – p + p + 1] = 0
(1 – p) = 0
p = 1
Therefore, given equation now becomes
x^{2} + x = 0
x(x + 1) = 0
x = 0, 1
The values of x and y besides y can satisfy the equation (x, y ∈ real numbers) x^{2}– xy + y^{2} – 4x – 4y + 16 = 0
For what value of a and b the equation x^{4} – 4x^{3} + ax^{2} + bx + 1 = 0 has four real positive roots ?
Since all roots are positive,
coeff. of x shld be negatuve i.e., b is negative
let roots be m(>0), n(>0), p(>0), q(>0)
m+n+p+q = 4
mn+np+pq+mp+mq+nq = a
mnp+mnq+npq+mpq = b
mnpq = 1
For the objective case,
put m=n=p=q =1
so a = 1+1+1+1+1+1 = 6
b = 1+1+1+1 =4
so, a = 6 and b = 4
If a, b are roots of the equation ax^{2} + bx + c = 0, then the value of a^{3} + b^{3} is
If z_{1} = 2 + i, z_{2} = 1 + 3i, then Re ( z_{1}  z_{2}) =
The numbers in the questions are not very clear.
z_{1} = 2 + i
z_{2} = 1+3i
z_{1} – z_{2} = (2 – 1) + i (1 – 3)
= 1 – 2i
If a, b are roots of the equation ax^{2} + bx + c = 0 then the equation whose roots are 2a + 3b and 3a + 2b is
If S is the set of all real x such that is positive, then S contains
If the roots of the equation x^{2}  5x + 16 = 0 are a, b and the roots of the equation x^{2} + px + q = 0 are (a^{2} + b^{2}) and , then
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