Solution 1
(i) Given: (x+5)(x-5)=24

(ii)
Given: ![]()

(iii)
Given: ![]()

or ![]()
(iv)
To find : x
Given quadratic equation is
…. (i)
One of the roots of (i) is
, so it satisfies (i)

So, the equation (i) becomes ![]()

Hence, the other root is
.
One root of the quadratic equation
is -3, find its other root.
Given quadratic equation is
…. (i)
One of the roots of (i) is -3, so it satisfies (i)

Hence, the other root is 2a.
If
and
;find the values of x.
So, the given quadratic equation becomes

Hence, the values of x are
and
.
Find the solution of the equation
; if
and
.
Given quadratic equation is
….. (i)
Also, given
and ![]()
and ![]()
So, the equation (i) becomes

Hence, the solution of given quadratic equation are
and
.
If m and n are roots of the equation
where x ≠ 0 and x ≠ 2; find m × n.
Given quadratic equation is ![]()

Since, m and n are roots of the equation, we have
and ![]()

Hence,
.
Solve, using formula :
Given quadratic equation is ![]()
Using quadratic formula,

⇒ x = a + 1 or x = -a - 2 = -(a + 2)
Solve the quadratic equation ![]()
(i) When
(integers)
(ii) When
(rational numbers)

(i) When
the equation
has no roots
(ii) When
the roots of
are
or ![]()
Find the value of m for which the equation
has real and equal roots.
Given quadratic equation is ![]()
The quadratic equation has real and equal roots if its discriminant is zero.

or ![]()
Find the values of m for which equation
has equal roots. Also, find the roots of the given equation.
Given quadratic equation is
…. (i)
The quadratic equation has equal roots if its discriminant is zero

When
, equation (i) becomes

When
, equation (i) becomes

∴ x = ![]()
Find the value of k for which equation
has real roots.
Given quadratic equation is
…. (i)
The quadratic equation has real roots if its discriminant is greater than or equal to zero

Hence, the given quadratic equation has real roots for
.
Find, using quadratic formula, the roots of the following quadratic equations, if they exist
(i) ![]()
(ii)
(i) Given quadratic equation is ![]()
D = b2 - 4ac =
= 25 - 24 = 1
Since D > 0, the roots of the given quadratic equation are real and distinct.
Using quadratic formula, we have

or ![]()
(ii) Given quadratic equation is ![]()
D = b2 - 4ac =
= 16 - 20 = - 4
Since D < 0, the roots of the given quadratic equation does not exist.
Solve :
(i)
and x > 0.
(ii)
and x < 0.
(i) Given quadratic equation is ![]()

or ![]()
But as x > 0, so x can't be negative.
Hence, x = 6.
(ii) Given quadratic equation is ![]()

or ![]()
But as x < 0, so x can't be positive.
Hence, ![]()
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