The quadratic equation is an expression having an order of degree 2 for
variables. First term of this sample quadratic equation is in the order
of two and the second term has order of degree 1 and the last term is
constant. If any of the term is not present in the quadratic equation
means that the term has a value 0.
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Samples of quadratic equation:
General form of quadratic equation is
ax2+bx+c.
As we saw the first degree of first term is 2, the degree of second term is 1 and third term is constant. Here the first two terms are jointly with a,b constants.
Solving a quadratic equation:
The quadratic equation can be solved by using the following formula.
x=`(-b +- sqrt(b^2 -4ac))/(2a)`
There is chance of getting two values as answers. Because of the plus,minus signs.
Solve the following sample quadratic equation of expression 5x2+2x+1
Solution:
Example: 2
Given quadratic equation is
= `(-2 +-sqrt(2^2-4.1.1))/(2.1)`
= `(-2 +- sqrt(4-4))/2`
=`(-2 +- sqrt(0))/2`
=`-2/2`
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Problem: 1
Solve the following sample quadratic equation of expression2x2+3x+2.
Answer: `(-3 +sqrt(7)i)/6` or `(-3 -sqrt(7)i)/6`
Problem: 2
Solve the following sample quadratic equation of expression 4x2+3x+1
Answer: `(-3 + sqrt(7)i)/8` or `(-3 - sqrt (7)i)/8`.
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Samples of quadratic equation:
Samples of quadratic equation:
General form of quadratic equation is
ax2+bx+c.
As we saw the first degree of first term is 2, the degree of second term is 1 and third term is constant. Here the first two terms are jointly with a,b constants.
Solving a quadratic equation:
The quadratic equation can be solved by using the following formula.
x=`(-b +- sqrt(b^2 -4ac))/(2a)`
There is chance of getting two values as answers. Because of the plus,minus signs.
Example problems for samples of quadratic equation:
Example: 1Solve the following sample quadratic equation of expression 5x2+2x+1
Solution:
Given quadratic equation is 5x2+2x+1
We know the general quadratic equation is ax2+bx+c.
Here a=5, b=2 and c=1.
The general formula for solving a quadratic equation is,
x=`(-b +- sqrt(b^2-4ac))/(2a)`
=`(-2 +- sqrt(2^2 -4.5.1))/(2.5)`
=`(-2 +- sqrt(4-20))/10`
=`(-2 +- sqrt (-16))/10`
We
can take square root of 16 but it has a negative sign. So we have to
consider it as imaginary part and insert i into the square root value.
=`(-2 +- 4i)/10`
= `(-1+- 2i)/5`
Answer of the above quadratic equation is `(-1+2i)/5` or `(-1 - 2i)/5`. Example: 2
Solve the following sample quadratic equation of expression x2+2x+1.
Solution:Given quadratic equation is
We know the general quadratic equation is ax2+bx+c.
The general formula for solving a quadratic equation is,
x=`(-b +- sqrt(b^2 -4ac))/(2a)`= `(-2 +-sqrt(2^2-4.1.1))/(2.1)`
= `(-2 +- sqrt(4-4))/2`
=`(-2 +- sqrt(0))/2`
=`-2/2`
= -1
Answer for the above quadratic equation is -1.I am planning to write more post on samacheer kalvi social book for 10th. Keep checking my blog.
Practice problems for samples quadratic equation:
Problem: 1
Solve the following sample quadratic equation of expression2x2+3x+2.
Answer: `(-3 +sqrt(7)i)/6` or `(-3 -sqrt(7)i)/6`
Problem: 2
Solve the following sample quadratic equation of expression 4x2+3x+1
Answer: `(-3 + sqrt(7)i)/8` or `(-3 - sqrt (7)i)/8`.
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