Notes on Quadratic equation | Grade 9 > Compulsory Maths > Algebra | KULLABS.COM

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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. The roots of the equations are :

$$x = \frac {-b \pm \sqrt {b^2 -4ac}}{2a}$$

#### Quadratic equation are of two types

A quadratic equation of the form ax2+ c = 0, where the middle term containing power 1 is missed is known as a pure quadratic equation.

e.g. x2 = 9 or, x2 - 9 = 0
i.e. ax2 + c = 0 ( a ≠ o, c = 0 )

The standard form of quardic equation is known as adfected quadratic equation.

e.g. x2 - 9x - 15 = 0
i.e. ax2 + 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

1. Factorization method
2. Completing square method
3. Using formula

a) Solving a quadratic equation by factorization method:

In this method, the quadratic equation ax2 + 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 ax2 + 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 ax2+ bx+ c = 0 can be obtained using a direct formula,

x = $$\frac{-b±\sqrt{b^2-4ac}}{2a}$$

'a' is the coefficient of x2

'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 ax2+ 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

1. Factorization method
2. Completing square method
3. Using formula
.

#### Click on the questions below to reveal the answers

Here,

x2 - 7x + 6 = 0

or, x2- 7x = -6

or,x2 - 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 x2 - 4x + 6 = 0 are 2.5 and 4.5.

Here,

x2 - 7x + 6 = 0

or, x2- 7x = -6

or,x2 - 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 x2 - 4x + 6 = 0 are 2.5 and 4.5.

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

Here, x2 - 25 = 0

or, x2 - (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.

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.

Here,

x2- 100 = 0

or, x2 - 102 = 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 = x2 - 100

= 102 - 100

= 0 = R.H.S proved, which is true

when x = -10,

L.H.S = x2 - 100

= -102 - 100

= 0 = R.H.S proved, which is true

∴ Hence, x = 10 and -10 both satisfy the equation.

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• ### ax2 + c = 0 is the form of ___________________ .

linear equation
factorization method
• ### ax2 + bx + c = 0 is the quadratic equation form of _____________________ .

completing square method
factorization method
• ### A second degree equation of one variable is known as ___________________ .

linear equation

method
formula
types
equation

x = 4 and 5
x = 7 and 4
x = 1 and -6
x = 2 and 3

3
-3
2
4

x = 5, 0
x = 0, 7
x = 2, 5
x = 2, 4

x = 2, 4
x = 2, 5
x = 0, 7
x = 5, 0
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