Algebraic Equations Definition is given as equations that are a sum of number of terms where each term is the multiplication of some constant term and positive integral power of some variables. Algebraic equations can be divided into following categories as:
Linear equations: these include all equations with degree 1 which means the maximum power of variables can be 1 only. For example: x+5=12.
Quadratic equations: these are the equations of degree 2. Graphically these are represented by a U shaped parabola. Mathematically they are represented as: ax2 + bx + c=0 where a 0.
Exponential equations: these are equations where a term includes variable as the power of constant value for example: ex + e =0. Or 2x + x =2 etc.
Logarithmic equations: these equations involves a logarithmic term which is given as: loga x. Here ‘a’ cannot be 1 and a >0, x>0. Let us see some methods of How to Solve Algebra Equations through some Algebraic Equations Examples now.
1) Linear equations: these equations are very easy to solve. Solution of the equation means to find out the value of variable that satisfies the given equation. Linear equations can be solved by isolating the variable of the equation. For ex:
2x+ 6 = 5.
to isolate x we can move 6 to RHS to get: 2x=5-6 = -1.
now a can be moved to RHS as: x= -1/2.
2) Quadratic equations: there are many methods to solve these equations like factoring, completing the square, quadratic formula and graphing. Here we will be quadratic formula to solve a quadratic equation.
The formula to solve a quadratic equation ax2 + bx + c=0 is given as:
x =
Example: solve 2x2 – 5x +2 =0.
By using above formula we have: a= 2, b =-5 and c =2.
x =
x =
x =
x=
So we get two solutions here as: x=(5+3)/4 and (5-3)/4 or 2, ½.
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3) Exponential equations: to solve these we first isolate exponential term and then take logarithm at both the sides of equation.
Example: e2x - 3ex+2=0.
This can be written as: (ex-1)(ex-2)=0
Equate both terms of product by zero:
ex - 1=0
ex =1
=> taking logarithm on both sides we get: ln(ex) = ln(1)
=> x (ln(e)) = 0
=>x.1=0. Hence x=0.
ex - 2=0
Similarly solving this we get x=ln(2)=0.69314 approx.
4) Logarithmic Algebra Equation is solved by taking exponential.
Example: ln(x)=3
eln(x) = e3
x=e3.
x=20.085537 approx.
Linear equations: these include all equations with degree 1 which means the maximum power of variables can be 1 only. For example: x+5=12.
Quadratic equations: these are the equations of degree 2. Graphically these are represented by a U shaped parabola. Mathematically they are represented as: ax2 + bx + c=0 where a 0.
Exponential equations: these are equations where a term includes variable as the power of constant value for example: ex + e =0. Or 2x + x =2 etc.
Logarithmic equations: these equations involves a logarithmic term which is given as: loga x. Here ‘a’ cannot be 1 and a >0, x>0. Let us see some methods of How to Solve Algebra Equations through some Algebraic Equations Examples now.
1) Linear equations: these equations are very easy to solve. Solution of the equation means to find out the value of variable that satisfies the given equation. Linear equations can be solved by isolating the variable of the equation. For ex:
2x+ 6 = 5.
to isolate x we can move 6 to RHS to get: 2x=5-6 = -1.
now a can be moved to RHS as: x= -1/2.
2) Quadratic equations: there are many methods to solve these equations like factoring, completing the square, quadratic formula and graphing. Here we will be quadratic formula to solve a quadratic equation.
The formula to solve a quadratic equation ax2 + bx + c=0 is given as:
x =
Example: solve 2x2 – 5x +2 =0.
By using above formula we have: a= 2, b =-5 and c =2.
x =
x =
x =
x=
So we get two solutions here as: x=(5+3)/4 and (5-3)/4 or 2, ½.
Please express your views of this topic quadratic formula discriminant by commenting on blog.
3) Exponential equations: to solve these we first isolate exponential term and then take logarithm at both the sides of equation.
Example: e2x - 3ex+2=0.
This can be written as: (ex-1)(ex-2)=0
Equate both terms of product by zero:
ex - 1=0
ex =1
=> taking logarithm on both sides we get: ln(ex) = ln(1)
=> x (ln(e)) = 0
=>x.1=0. Hence x=0.
ex - 2=0
Similarly solving this we get x=ln(2)=0.69314 approx.
4) Logarithmic Algebra Equation is solved by taking exponential.
Example: ln(x)=3
eln(x) = e3
x=e3.
x=20.085537 approx.
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