Subject: Compulsory Maths
An equation like ax2+ bx + c = 0 where a≠ 0, which contains only one variable and '2' as its highest power is called a quadratic equation. There are two values of the variable in any quadratic equation.
An equation like ax^{2}+ bx + c = 0 where a ≠ 0, which contains only one variable and '2' as its highest power is called a quadratic equation. There are two values of the variable in any quadratic equation. The roots of the equations are :
$$ x = \frac {-b \pm \sqrt {b^2 -4ac}}{2a} $$
i. Pure quadratic equation:
A quadratic equation of the form ax^{2}+ c = 0, where the middle term containing power 1 is missed is known as a pure quadratic equation.
e.g. x^{2} = 9 or, x^{2} - 9 = 0
i.e. ax^{2} + c = 0 ( a ≠ o, c = 0 )
ii. Adfected quadratic equation:
The standard form of quardic equation is known as adfected quadratic equation.
ax^{2} + bx + c = 0 is adfected quadratic equation.
e.g. x^{2} - 9x - 15 = 0
i.e. ax^{2} + bx + c = 0 ( a ≠ o, b ≠ o)
Quadratic equation is a second-degree equation of one variable. So we get two solutions of the variable contained by this equation. The solution of a quadratic equation is known as the roots. Hence, a quadratic equation has two roots. The solution of roots which are obtained from the quadratic equation should satisfy the equation. There are three major methods for solving quadratic equation, which are
a) Solving a quadratic equation by factorization method:
In this method, the quadratic equation ax^{2} + bx+ c = 0 is factorized and expressed as the product of two linear factors. Again, the two linear factors are also solved to get the solution of the equation. the roots that is obtained from the equation should satisfy the given quadratic equation.
b) Solving a quadratic equation by completing square method:
We transpose the constant term (c) to the R.H.S and the remaining parts ax^{2} + bx is expressed in the perfect square expression.
c) Solving a quadratic equation by using formula:
In this method, we have obtained two roots as \(\frac{-b+\sqrt{b^2-4ac}}{2a}\) and \(\frac{-b-\sqrt{b^2-4ac}}{2a}\)
the two roots of the quadratic equation ax^{2}+ bx+ c = 0 can be obtained using a direct formula,
x = \(\frac{-b±\sqrt{b^2-4ac}}{2a}\)
'a' is the coefficient of x^{2}
'b' is the coefficient of x and
'c' is the constant term.
While solving any quadratic equation using the formula, first we simplify and bring the equation in the simplest form. Then we compare with ax^{2}+ bx+ c = 0 and find a, b and c. Finally, we use x = \(\frac{-b±\sqrt{b^2-4ac}}{2a}\) and find two roots of the given quadratic equation.
There are three major methods for solving quadratic equation, which are
Solve the quadratic equation x^{2 }- 7x + 6 = 0 by completing square method.
Here,
x^{2} - 7x + 6 = 0
or, x^{2}- 7x = -6
or,x^{2} - 2.x.(7/2) + (7/2)^{2}= (7/2)^{2} - 6
or,(x - 7/2)^{2} =14/2 - 6
or,(x - 7/2)^{2} = 7 - 6
or, (x - 7/2)^{2} = (±1)^{2}
Avoiding square on both sides, we get
x - 7/2 =±1
or, x = 7/2±1
Now, taking +ve sign, we get
x = 7/2 + 1 = 7+2/2 = 9/2 = 4.5
Now, taking -ve sign, we get
x = 7/2 - 1 = 7-2/2 = 5/2 = 2.5
∴ x = 4.5 or 2.5
Hence, the two roots of the quadratic equation x^{2} - 4x + 6 = 0 are 2.5 and 4.5.
Solve the quadratic equation x^{2 }- 7x + 6 = 0 by completing square method.
Here,
x^{2} - 7x + 6 = 0
or, x^{2}- 7x = -6
or,x^{2} - 2.x.(7/2) + (7/2)^{2}= (7/2)^{2} - 6
or,(x - 7/2)^{2} =14/2 - 6
or,(x - 7/2)^{2} = 7 - 6
or, (x - 7/2)^{2} = (±1)^{2}
Avoiding square on both sides, we get
x - 7/2 =±1
or, x = 7/2±1
Now, taking +ve sign, we get
x = 7/2 + 1 = 7+2/2 = 9/2 = 4.5
Now, taking -ve sign, we get
x = 7/2 - 1 = 7-2/2 = 5/2 = 2.5
∴ x = 4.5 or 2.5
Hence, the two roots of the quadratic equation x^{2} - 4x + 6 = 0 are 2.5 and 4.5.
Solve x^{2} - 8x + 12 = 0 quadratic equation by factorization method. Also verify it.
Here, x^{2} - 8x + 12 =0
or, x^{2} - (6 + 2)x + 12 = 0
or, x^{2} - 6x - 2x + 12 = 0
or, x (x - 6) -2 (x - 6) = 0
or, (x - 6) (x - 2) = 0
∴ x - 6 = 0
or, x = 6
Again, x - 2 = 0
or, x = 2
∴ x = 2 and 6
Now verification,
When x = 2
L.H.S. = x^{2}- 8x + 12
= 2^{2} - 8*2 + 12
= 4 - 16 +12
= 0
= R.H.S proved, which is true
Again when x = 6,
L.H.S. = x^{2} - 8x + 12
= 6^{2 -} 8*6 + 12
= 36 - 48 + 12
= 0
= R.H.S. proved, which is true
∴ x = 2 and 6 satisfy the equation. Hence the two roots of the quadratic equation of x^{2}- 8x +12 = 0 are 2 and 6.
Solve x^{2} - 8x + 12 = 0 quadratic equation by using formula and verify it.
Here, comparing the equation x^{2} - 8x + 12 = 0 with ax^{2} - bx + c = 0.
We get, a = 1, b = -8 and c = 12.
By using formula,
x = \(\frac{(-b) ±\sqrt(b^2 - 4ac)}{2a}\)
= \(\frac{(-(-8)) ±\sqrt(-8^2 - 4*1*12)}{2*1}\)
= \(\frac{(8) ±\sqrt(64 - 48)}{2}\)
=\(\frac{(8) ±\sqrt(16)}{2}\)
=\(\frac{8± 4 }{2}\)
Now, taking +ve sign, we get
x =\(\frac{8 + 4}{2}\)
= 6
Again, taking -ve sign, we get
x =\(\frac{8 - 4}{2}\)
= 2
∴ 2 or 6
Now verification,
When x = 2,
L.H.S. = x^{2} - 8x + 12
= 2^{2} - 8*2 + 12
= 4 - 16 +12
= 0 = R.H.S. proved, which is true
When x = 6,
L.H.S. = x^{2} - 8x + 12
= 6^{2} - 8*6 + 12
= 36 - 48 + 12
= 0 = R.H.S. proved, which is true
∴ x = 2 and 6 both satisfy the equation.
Solve the quadratic equation x^{2} - 5x + 6 = 0 by completing square method and verify it.
Here, x^{2} - 5x + 20 = 0
or, x^{2} - 5x = -6
or, (x)^{2}- 5x = -6
or, (x)^{2} - 2.x.(5/2) + (5/2)^{2} = (5/2)^{2} -6
or, (x - 5/2)^{2} = 25/4 - 6
or, (x - 5/2)^{2} = 25 - 24/4
or, (x - 5/2)^{2} = 1/4
or, (x - 5/2)^{2} = (± 1/2)^{2}
Avoiding squares on both sides,
x - 5/2 =±1/2
x = 5/2± 1/2
Now, taking +ve sign, we get
x = 5+1/2 = 3
Taking -ve sign, we get
x = 5-1/2 = 2
∴ x = 3 or 2
Now verification,
When x = 2,
L.H.S. = x^{2}- 5x + 6
= 2^{2} - 5*2 + 6
= 4 - 10 + 6
= 0 = R.H.S. proved, which is true.
When x = 3,
L.H.S. = x^{2} - 5x + 6
= 3^{2} - 5*3 + 6
= 9 - 15 + 6
= 0 = R.H.S proved, which is true
∴ x = 2 and 3 both satisfy the equation.
Solve the quadratic equation (x + 1) (x + 4) = 0.
Here, (x - 1) (x + 4) = 0
Either, x - 1 = 0.............(i)
or, x + 4 = 0............(ii)
Now, from equation (i),
x - 1 = 0
or, x = 1
Again, from equation (ii),
x + 4 = 0
or, x = -4
∴ x = 1 and -4.
Solve the quadratic equation (x -3) (x + 2) = 0.
Here, (x - 3 ) (x + 2) = 0
Either, x - 3 = 0...............(i)
Or, x + 2 = 0...............(ii)
Now, from equation (i),
x - 3 = 0
or, x = 3
Now, from equation (ii),
x + 2 = 0
or, x = -2
∴ x = 3 or -2
Solve: x^{2} - 49 = 0
Here, x^{2} - 49 = 0
or, x^{2} - (7)^{2} = 0
or, (x + 7) (x - 7) = 0
Either, x + 7 = 0.................(i)
Or, x - 7 = 0.....................(ii)
Now from equation (i),
or, x + 7 = 0
or, x = -7
Now from equation (ii),
or, x - 7 = 0
or, x = 7
∴ x =±7 both satisfy the equation.
Solve the quadratic equation 3x^{2} - 5x - 2 =0 by using formula and verify it.
Here, comparing the equation 3x^{2}- 5x - 2 = 0 with ax^{2} + bx + c =0.
We get, a = 3, b = -5 and c = -2
By usinf formula,
x = \(\frac{(-b) ±\sqrt(b^2 - 4ac)}{2a}\)
= \(\frac{(-(-5)) ±\sqrt(-5^2 - 4*3*-2)}{2*3}\)
= \(\frac{(5) ±\sqrt(25 + 24)}{6}\)
= \(\frac{5 ±\sqrt(49)}{6}\)
= \(\frac{5 ± 7}{6}\)
Now, taking +ve sign,we get
x = \(\frac{5 + 7}{6}\)
= 2
Again, taking -ve sign we get,
x = \(\frac{5 - 7}{6}\)
= \(\frac {-1} {3}\)
∴ x = 2 or\(\frac {-1} {3}\)
Now verification,
When x = 2
L.H.S. = 3x^{2} - 5x - 2
= 3*2^{2} - 5 * 2 - 2
= 12 - 10 - 2
= 0 = R.H.S, which is true
When x =\(\frac {-1} {3}\)
L.H.S. x = 3x^{2} - 5x - 2
= 3*(\(\frac {-1} {3}\))^{2}- 5\(\frac {-1} {3}\) - 2
= 1 +\(\frac {5} {3}\) - 2
= 0 = R.H.S, which is true
∴ x = 2 and\(\frac {-1} {3}\) both satisfy the equation.
Solve x^{2} - 7x + 12 = 0.
Here, x = x^{2} - 7x + 12 = 0
or, x^{2} - (4 + 3)x + 12 = 0
or, x^{2} -4x -3x + 12 = 0
or, x(x - 4) -3(x - 4) = 0
or, (x - 4) (x - 3) = 0
Either, x - 4 = 0
∴x = 4
Or, x - 3 = 0
∴x = 3
Now verification,
When, x = 4
L.H.S. = x^{2} - 7x + 12 = 0
= 4^{2} - 7*4 + 12
= 16 - 28 + 12
= 0 = R.H.S, which is true
When x = 3
L.H.S = x^{2} - 7x + 12
= 3^{2} - 7*3 + 12
= 9 - 21 + 12
= 0 = R.H.S, which is true
∴ x = 4 and 3 both satisfy the equation.
Solve: x^{2} - 25
Here, x^{2} - 25 = 0
or, x^{2} - (5)^{2} = 0
or, (x + 5) (x - 5) = 0
Either, x + 5 = 0.................(i)
Or, x - 5 = 0.....................(ii)
Now from equation (i),
or, x + 5 = 0
or, x = -5
Now from equation (ii),
or, x - 5 = 0
or, x = 5
∴ x =±5 both satisfy the equation.
Solve x^{2} - 15x + 36 = 0 quadratic equation by factorization method. Also verify it.
Here, x^{2} - 15x + 36 =0
or, x^{2} - (12 + 3)x + 36 = 0
or, x^{2} - 12x - 3x + 36 = 0
or, x (x - 12) -3 (x - 12) = 0
or, (x - 12) (x - 3) = 0
∴ x - 12 = 0
or, x = 12
Again, x - 3 = 0
or, x = 3
∴ x = 12 and 3
Now verification,
When x = 12
L.H.S. = x^{2}- 15x + 36
= 12^{2} - 8*12 + 36
= 144 - 180 + 36
= 0
= R.H.S proved, which is true
Again when x = 3,
L.H.S. = x^{2} - 8x + 12
= 3^{2 -} 8*3 + 36
= 9 - 45 + 36
= 0
= R.H.S. proved, which is true
∴ x = 3 and 12 satisfy the equation. Hence the two roots of the quadratic equation of x^{2}- 8x +12 = 0 are 3 and 12.
Solve x^{2} - 15x + 36 = 0 quadratic equation by using formula and verify it.
Here, comparing the equation x^{2} - 15x + 36 = 0 with ax^{2} - bx + c = 0.
We get, a = 1, b = -15 and c = 36.
By using formula,
x = \(\frac{(-b) ±\sqrt(b^2 - 4ac)}{2a}\)
= \(\frac{(-(-15)) ±\sqrt(-15^2 - 4*1*36)}{2*1}\)
= \(\frac{(15) ±\sqrt(225 - 144)}{2}\)
=\(\frac{(15) ±\sqrt(81)}{2}\)
=\(\frac{15 ± 9 }{2}\)
Now, taking +ve sign, we get
x =\(\frac{15 + 9}{2}\)
= 12
Again, taking -ve sign, we get
x =\(\frac{15 - 9}{2}\)
= 3
∴ 3 or 12
Now verification,
When x = 3,
L.H.S. = x^{2} - 15x + 36
= 3^{2} - 15*3 + 36
= 9 - 45 +36
= 0 = R.H.S. proved, which is true
When x = 12,
L.H.S. = x^{2} - 15x + 36
= 12^{2} - 15*12 + 36
= 144 - 180 + 36
= 0 = R.H.S. proved, which is true
∴ x = 3 and 12 both satisfy the equation.
Solve the quadratic equation (2x - 5) (x - 3) and verify it.
Here, (2x - 5 ) (x + 3) = 0
Either, 2x - 5 = 0...............(i)
Or, x + 3 = 0...............(ii)
Now, from equation (i),
2x - 5 = 0
or, x = 5/2
Now, from equation (ii),
x + 3 = 0
or, x = -3
∴ x = 5/2 or -3
Now, verification
when x = 5/2
L.H.S = (2x - 5) (x +3)
= (2*5/2 - 5) (5/2 + 3)
= 0 = R.H.S proved, which is true
when x = -3
L.H.S = (2x - 5) (x + 3)
= (2*-3 - 5) (-3 + 3)
= 0 = R.H.S proved, which is true
∴ 5/2 and -3 both satisfy the equation.
Solve: x^{2} - 100 = 0
Here,
x^{2}- 100 = 0
or, x^{2} - 10^{2} = 0
or, (x - 10) (x + 10) = 0
Either, x - 10 = 0..............(i)
Or, x + 10 = 0...............(ii)
Now,
x - 10 = 0
or, x = 10
Again,
x + 10 = 0
or, x = -10
∴x =±10
Now, verification
when x = 10
L.H.S = x^{2} - 100
= 10^{2} - 100
= 0 = R.H.S proved, which is true
when x = -10,
L.H.S = x^{2} - 100
= -10^{2} - 100
= 0 = R.H.S proved, which is true
^{}∴ Hence, x = 10 and -10 both satisfy the equation.
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