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In order to solve a quadratic equation of the form ax2 + bx + c, we first need to calculate the discriminant with the help of the formula D = b2 β 4ac.
The solution of the quadratic equation ax2 + bx + c= 0 is given by x = [-b Β± β b2 β 4ac] / 2a
If Ξ± and Ξ² are the roots of the quadratic equation ax2 + bx + c = 0, then we have the following results for the sum and product of roots:
Ξ± + Ξ² = -b/a
Ξ±.Ξ² = c/a
Ξ± β Ξ² = βD/a
It is not possible for a quadratic equation to have three different roots and if in any case it happens, then the equation becomes an identity.
Nature of Roots:
Consider an equation ax2 + bx + c = 0, where a, b and c β R and a β 0, then we have the following cases:
D > 0 iff the roots are real and distinct i.e. the roots are unequal
D = 0 iff the roots are real and coincident i.e. equal
D < 0 iffthe roots are imaginary
The imaginary roots always occur in pairs i.e. if a+ib is one root of a quadratic equation, then the other root must be the conjugate i.e. a-ib, where a, b β R and i = β-1.
Consider an equation ax2 + bx + c = 0, where a, b and c βQ and a β 0, then
If D > 0 and is also a perfect square then the roots are rational and unequal.
If Ξ± = p + βq is a root of the equation, where βpβ is rational and βq is a surd, then the other root must be the conjugate of it i.e. Ξ² = p - βq and vice versa.
If the roots of the quadratic equation are known, then the quadratic equation may be constructed with the help of the formula
x2 β (Sum of roots)x + (Product of roots) = 0.
So if Ξ± and Ξ² are the roots of equation then the quadratic equation is
x2 β (Ξ± + Ξ²)x + Ξ± Ξ² = 0
For the quadratic expressiony = ax2 + bx + c, where a, b, c β R and a β 0, then the graph between x and y is always a parabola.
If a > 0, then the shape of the parabola is concave upwards
If a < 0, then the shape of the parabola is concave upwards
Inequalities of the form P(x)/ Q(x) > 0 can be easily solved by the method of intervals of number line rule.
The maximum and minimum values of the expression y = ax2 + bx + c occur at the point x = -b/2a depending on whether a > 0 or a< 0.
y β[(4ac-b2) / 4a, β] if a > 0
If a < 0, then y β [-β, (4ac-b2) / 4a]
The quadratic function of the form f(x, y) = ax2+by2 + 2hxy + 2gx + 2fy + c = 0 can be resolved into two linear factors provided it satisfies the following condition: abc + 2fgh βaf2 β bg2 β ch2 = 0
In general, if Ξ±1,Ξ±2, Ξ±3, β¦β¦ ,Ξ±n are the roots of the equation
f(x) = a0xn +a1xn-1 + a2xn-2 + β¦β¦. + an-1x + an, then
1.Σα1 = - a1/a0
2.Ξ£ Ξ±1Ξ±2 = a2/a0
3.Ξ£ Ξ±1Ξ±2Ξ±3 = - a3/a0
β¦β¦β¦ β¦β¦β¦.
Ξ£ Ξ±1Ξ±2Ξ±3 β¦β¦Ξ±n= (-1)n an/a0
Every equation of nth degree has exactly n roots (n β₯1) and if it has more than n roots then the equation becomes an identity.
If there are two real numbers βaβ and βbβ such that f(a) and f(b) are of opposite signs, then f(x) = 0 must have at least one real root between βaβ and βbβ.
Every equation f(x) = 0 of odd degree has at least one real root of a sign opposite to that of its last term.
In order to solve a quadratic equation of the form ax2 + bx + c, we first need to calculate the discriminant with the help of the formula D = b2 β 4ac.
The solution of the quadratic equation ax2 + bx + c= 0 is given by x = [-b Β± β b2 β 4ac] / 2a
If Ξ± and Ξ² are the roots of the quadratic equation ax2 + bx + c = 0, then we have the following results for the sum and product of roots:
Ξ± + Ξ² = -b/a
Ξ±.Ξ² = c/a
Ξ± β Ξ² = βD/a
It is not possible for a quadratic equation to have three different roots and if in any case it happens, then the equation becomes an identity.
Nature of Roots:
Consider an equation ax2 + bx + c = 0, where a, b and c β R and a β 0, then we have the following cases:
D > 0 iff the roots are real and distinct i.e. the roots are unequal
D = 0 iff the roots are real and coincident i.e. equal
D < 0 iffthe roots are imaginary
The imaginary roots always occur in pairs i.e. if a+ib is one root of a quadratic equation, then the other root must be the conjugate i.e. a-ib, where a, b β R and i = β-1.
Consider an equation ax2 + bx + c = 0, where a, b and c βQ and a β 0, then
If D > 0 and is also a perfect square then the roots are rational and unequal.
If Ξ± = p + βq is a root of the equation, where βpβ is rational and βq is a surd, then the other root must be the conjugate of it i.e. Ξ² = p - βq and vice versa.
If the roots of the quadratic equation are known, then the quadratic equation may be constructed with the help of the formula
x2 β (Sum of roots)x + (Product of roots) = 0.
So if Ξ± and Ξ² are the roots of equation then the quadratic equation is
x2 β (Ξ± + Ξ²)x + Ξ± Ξ² = 0
For the quadratic expressiony = ax2 + bx + c, where a, b, c β R and a β 0, then the graph between x and y is always a parabola.
If a > 0, then the shape of the parabola is concave upwards
If a < 0, then the shape of the parabola is concave upwards
Inequalities of the form P(x)/ Q(x) > 0 can be easily solved by the method of intervals of number line rule.
The maximum and minimum values of the expression y = ax2 + bx + c occur at the point x = -b/2a depending on whether a > 0 or a< 0.
y β[(4ac-b2) / 4a, β] if a > 0
If a < 0, then y β [-β, (4ac-b2) / 4a]
The quadratic function of the form f(x, y) = ax2+by2 + 2hxy + 2gx + 2fy + c = 0 can be resolved into two linear factors provided it satisfies the following condition: abc + 2fgh βaf2 β bg2 β ch2 = 0
In general, if Ξ±1,Ξ±2, Ξ±3, β¦β¦ ,Ξ±n are the roots of the equation
f(x) = a0xn +a1xn-1 + a2xn-2 + β¦β¦. + an-1x + an, then
1.Σα1 = - a1/a0
2.Ξ£ Ξ±1Ξ±2 = a2/a0
3.Ξ£ Ξ±1Ξ±2Ξ±3 = - a3/a0
β¦β¦β¦ β¦β¦β¦.
Ξ£ Ξ±1Ξ±2Ξ±3 β¦β¦Ξ±n= (-1)n an/a0
Every equation of nth degree has exactly n roots (n β₯1) and if it has more than n roots then the equation becomes an identity.
If there are two real numbers βaβ and βbβ such that f(a) and f(b) are of opposite signs, then f(x) = 0 must have at least one real root between βaβ and βbβ.
Every equation f(x) = 0 of odd degree has at least one real root of a sign opposite to that of its last term.