An equation with one variable, in which the highest power of the variable is two is called a quadratic equation.
For example, ( i ) 3x2 + 5 x - 8 = 0
(ii) 2y2 - 48 = 0
(iii) 6x2 + 5x = 0
(iv ) y2 = 4 are all quadratic equations. Let us do some activity or problems using quadratic equations.
Activities of Quadratic Equations:
Activity 1 :
Solve the quadratic equation 2x2 - 7x = 39
2x2 - 7x - 39 - 0
`=>` 2x2 - 13x + 6x - 39 = 0 ( factorising the left hand side )
`=>` x ( 2x - 13 ) + 3 (2 x - 13 ) = 0
`=>` ( 2x - 13 ) ( x + 3 ) = 0
2x - 13 = 0 or x + 3 = 0
`rArr` x = `13/2` or x = -3
Hence the quadratic equation is solved by factorisation method.
Activity 2 :
Find the quadratic equation whose solution set is { -2, 3 }
Since solution set is { -2, 3}
we have x = -2 or x = 3
x + 2 = 0 or x - 3 = 0
`rArr` ( x + 2 ) ( x - 3 ) = 0
`rArr` x2 - 3x + 2x - 6 = 0
`rArr` x2 - x - 6 = 0 is the required quadratic equation.
Activity of Quadratic Equations(continued):
Activity 3 :
Solve the quadratic equation 5x2 - 2x - 3 = 0 using the formula.
The roots of the standard quadratic equation ax2 + bx + c = 0 where a`!=` 0, are given by the formula
x = `( -b stackrel(+)(-) sqrt ( b^2 - 4ac )) / ( 2a)`
Comparing 5x2 - 2x - 3 = 0 with ax2 + bx + c = 0 we get a = 5, b = -2 and c = -3.
so, x = `(2 stackrel(+)(-) sqrt((-2)^2 - 4. 5. (-3))/(2.5))`
= `(2 stackrel( +)(-) sqrt ( 64)) / ( 10)`
= `(2 stackrel(+)(-) 8)/10`
= `(2-8)/10` = 1 and `-3/5`
Hence 1 and `(-3)/5` are the roots of the given quadratic equation.
Activity 4 :
Solve the equation 2x4 - 5x2 + 3 + 0 which is reducible to quadratic equation.
Let x2 = y
Then, 2x4 - 5x2 + 3 = 0 `rArr` 2y2 - 5y + 3 + 0
`rArr` ( y - 1 ) ( 2y - 3 ) = 0
`rArr` y = 1 or y = `3/2`
When y = 1, x2 = 1 `rArr` x = 1 or -1
When y = `3/2` x2 = `3/2` `rArr` x = `sqrt(3/2)` or `-sqrt(3/2)`
Hence the fourth degree equation is solved using the quadratic equation technique.
For example, ( i ) 3x2 + 5 x - 8 = 0
(ii) 2y2 - 48 = 0
(iii) 6x2 + 5x = 0
(iv ) y2 = 4 are all quadratic equations. Let us do some activity or problems using quadratic equations.
Activities of Quadratic Equations:
Activity 1 :
Solve the quadratic equation 2x2 - 7x = 39
2x2 - 7x - 39 - 0
`=>` 2x2 - 13x + 6x - 39 = 0 ( factorising the left hand side )
`=>` x ( 2x - 13 ) + 3 (2 x - 13 ) = 0
`=>` ( 2x - 13 ) ( x + 3 ) = 0
2x - 13 = 0 or x + 3 = 0
`rArr` x = `13/2` or x = -3
Hence the quadratic equation is solved by factorisation method.
Activity 2 :
Find the quadratic equation whose solution set is { -2, 3 }
Since solution set is { -2, 3}
we have x = -2 or x = 3
x + 2 = 0 or x - 3 = 0
`rArr` ( x + 2 ) ( x - 3 ) = 0
`rArr` x2 - 3x + 2x - 6 = 0
`rArr` x2 - x - 6 = 0 is the required quadratic equation.
Activity of Quadratic Equations(continued):
Activity 3 :
Solve the quadratic equation 5x2 - 2x - 3 = 0 using the formula.
The roots of the standard quadratic equation ax2 + bx + c = 0 where a`!=` 0, are given by the formula
x = `( -b stackrel(+)(-) sqrt ( b^2 - 4ac )) / ( 2a)`
Comparing 5x2 - 2x - 3 = 0 with ax2 + bx + c = 0 we get a = 5, b = -2 and c = -3.
so, x = `(2 stackrel(+)(-) sqrt((-2)^2 - 4. 5. (-3))/(2.5))`
= `(2 stackrel( +)(-) sqrt ( 64)) / ( 10)`
= `(2 stackrel(+)(-) 8)/10`
= `(2-8)/10` = 1 and `-3/5`
Hence 1 and `(-3)/5` are the roots of the given quadratic equation.
Activity 4 :
Solve the equation 2x4 - 5x2 + 3 + 0 which is reducible to quadratic equation.
Let x2 = y
Then, 2x4 - 5x2 + 3 = 0 `rArr` 2y2 - 5y + 3 + 0
`rArr` ( y - 1 ) ( 2y - 3 ) = 0
`rArr` y = 1 or y = `3/2`
When y = 1, x2 = 1 `rArr` x = 1 or -1
When y = `3/2` x2 = `3/2` `rArr` x = `sqrt(3/2)` or `-sqrt(3/2)`
Hence the fourth degree equation is solved using the quadratic equation technique.
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