Revision: Algebra1 >> Quadratic Equations Applied Mathematics ISC (Commerce) Class 11 CISCE

An equation with one variable, in which the highest power of the variable is two, is known as a quadratic equation.

Standard Form:

 ax² + bx + c = 0,  a ≠ 0

For example :

(i) 3x² + 4x + 7 = 0

(ii) 4x² + 5x = 0 

A value of the variable which satisfies the equation is called a root (solution).

If substituting a value of x makes the polynomial zero, that value is a root.

• A number α is called a root of ax² + bx + c = 0, if aα² + bα + c = 0

Given the roots of the equation (q – r)x² + (r – p)x + (p – q) = 0 are equal.

∴ Discriminant (D) = 0

⇒ b² – 4ac = 0

⇒ (r – p)² – 4 × (q – r) × (p – q) = 0

⇒ r² + p² – 2pr – 4[qp – q² – rp + qr] = 0

⇒ r² + p² – 2pr – 4qp + 4q² + 4rp – 4qr = 0

⇒ r² + p² + 2pr – 4qp – 4qr + 4q² = 0

⇒ (p + r)² – 4q(p + r) + 4q² = 0

Let (p + r) = y

⇒ y² – 4qy + 4q² = 0

⇒ (y – 2q)² = 0

⇒ y – 2q = 0

⇒ y = 2q

⇒ p + r = 2q

Hence proved.

The given quadric equation is (a² + b²)x² − 2(ac + bd)x + (c² + d²) = 0, and roots are real

Then prove that `a/b=c/d`.

Here,

a = (a² + b²), b = -2 (ac + bd) and c = (c² + d²)

As we know that D = b² - 4ac

Putting the value of a = (a² + b²), b = -2 (ac + bd) and c = (c² + d²)

D = b² - 4ac

= {-2(ac + bd)}² - 4 × (a² + b²) × (c² + d²)

= 4(a²c² + 2abcd + b² + d²) - 4(a²c² + a²d² + b²c² + b²d²)

= 4a²c² + 8abcd + 4b²d² - 4a²c² - 4a²d² - 4b²c² - 4b²d²

= -4a²d² - 4b²c² + 8abcd

= -4(a²d² + b²c² - 2abcd)

The given equation will have real roots, if D = 0

-4(a²d² + b²c² - 2abcd) = 0

a²d² + b²c² - 2abcd = 0

(ad)² + (bc)² - 2(ad)(bc) = 0

(ad - bc)² = 0

Square root both sides we get,

ad - bc = 0

ad = bc

`a/b=c/d`

Hence `a/b=c/d`.

The given equation is (a² + b²)x² + 2(ac + bd)x + (c² + d²) = 0

We know, D = b² – 4ac

Thus,

D = [2(ac + bd)²] – 4(a² + b²)(c² + d²)

= [4(a²c² + b²d² + 2abcd)] – 4(a² + b²)(c² + d²)

= 4[(a²c² + b²d² + 2abcd) – (a²c² + a²d² + b²c² + b²d²)]

= 4[a²c² + b²d² + 2abcd – a²c² – a²d² – b²c² – b²d²]

= 4[2abcd – b²c² – a²d²]

= –4[a²d² + b²c² – 2abcd]

= –4[ad – bc]²

But we know that ad ≠ bc

Therefore, 

(ad – bc) ≠ 0

⇒ (ad – bc)² > 0

⇒ –4(ad – bc)² < 0

⇒ D < 0

Hence, the given equation has no real roots.

The given quadric equation is (b − c) x² + (c − a) x + (a − b) = 0, and roots are real

Then prove that 2b = a + c

Here,

a = (b − c), b = (c − a) and c = (a − b)

As we know that D = b² − 4ac

Putting the value of a = (b − c), b = (c − a) and c = (a − b)

D = b² − 4ac

= (c − a)² − 4 × (b − c) × (a − b)

= c² − 2ca + a² − 4 (ab − b² − ca + bc)

= c² − 2ca + a² − 4ab + 4b² + 4ca − 4bc

= c² + 2ca + a² − 4ab + 4b² − 4bc

= a² + 4b² + c² + 2ca − 4ab − 4bc

As we know that (a² + 4b² + c² + 2ca − 4ab − 4bc) = (a + c − 2b)²

D = (a + c − 2b)²

The given equation will have real roots, if D = 0

(a + c − 2b)² = 0

Square root both side we get

`sqrt((a + c - 2b)^2)=0`

a + c − 2b = 0

a + c = 2b

Hence 2b = a + c.

The given equations are

ax² + 2bx + c = 0             ............ (1)

`bx^2-2sqrt(ac)x+b = 0` ............. (2)

Roots are simultaneously real

Then prove that b² = ac

Let D₁ and D₂ be the discriminants of equation (1) and (2) respectively,

Then,

D₁ = (2b)² - 4ac

= 4b² - 4ac

And

`D_2=(-2sqrt(ac))^2-4xxbxxb`

= 4ac - 4b²

Both the given equation will have real roots, if D₁ ≥ 0 and D₂ ≥ 0

4b² - 4ac ≥ 0

4b² ≥ 4ac

b² ≥ ac                ............... (3)

4ac - 4b² ≥ 0

4ac ≥ 4b²

ac ≥ b²                          ................... (4)

From equations (3) and (4) we get

b² = ac

Hence, b² = ac.

The given equation \[\left(1 + m^2 \right) x^2 + 2mcx + \left(c^2 - a^2 \right) = 0\], has equal roots

Then prove that`c^2 = (1 + m^2)`.

Here,

`a = (1 + m^2), b = 2mc and c = (c^2 - a^2)`

As we know that `D = b^2 - 4ac`

Putting the value of `a = (1 + m^2), b = 2mc and c = (c^2 -  a^2)`

`D = b^2 - 4ac`

` = {2mc}^2 - 4xx (1 +m^2) xx (c^2 - a^2)`

` = 4 (m^2 c^2) - 4(c^2 -a^2 + m^2c^2 - m^2 a^2)`

` = 4m^2c^2 - 4c^2 + 4a^2 - 4m^2 c^2 + 4m^2a^2`

` = 4a^2 + 4m^2 a^2 = 4c^2`

The given equation will have real roots, if D  = 0

 `4a^2 + 4m^2 a^2 - 4c^2 = 0`

`4a^2 + 4m^2a^2 = 4c^2`

`4a^2 + (1 + m^2 ) = 4c^2`

`a^2 (1 +m^2) = c^2`

Hence, `c^2 = a^2 (1 + m^2)`.

The quadratic equation is (x - a)(x - b) + (x - b)(x - c) + (x - c)(x - a) = 0

Here,

After simplifying the equation

x² - (a + b)x ab + x² - (b + c)x + bc + x² - (c + a)x + ca

3x² - 2(a + b + c)x + (ab + bc + ca) = 0

a = 3, b = - 2(a + b + c) and c = (ab + bc + ca)

As we know that D = b² - 4ac

Putting the value of a = 3, b = - 2(a + b + c) and c = (ab + bc + ca)

D = {- 2(a + b + c)}² - 4 × (3) × (ab + bc + ca)

= 4(a² + b² + c² + 2ab + 2bc+ 2ca) - 12(ab + bc + ca)

= 4(a² + b² + c² + 2ab + 2bc+ 2ca) - 12ab - 12bc - 12ca

= 4(a² + b² + c² + 2ab + 2bc+ 2ca - 3ab - 3bc - 3ca)

= 4(a² + b² + c² - ab - bc - ca)

D = 4(a² + b² + c² - ab - bc - ca)

= 2[2a² + 2b² + 2c² - 2ab - 2ac - 2bc]

= 2[(a - b)² + (b - c)² + (c - a)²]

Since, D > 0. So the solutions are real

Let a = b = c

Then

D = 4(a² + b² + c² - ab - bc - ca)

= 4(a² + b² + c² - aa - bb - cc)

= 4(a² + b² + c² - a² - b² - c²)

= 4 × 0

Thus, the value of D = 0.

Therefore, the roots of the given equation are real and but they are equal only when, a = b = c.

Hence proved.

• A quadratic equation has degree 2.
• Its standard form is \[ ax^2 + bx + c = 0, \] where \[ a \neq 0. \]
• A root is a value of \[ x \] that satisfies the equation.
• A quadratic equation has at most two roots.

D = b² – 4ac 

Endpoints:

• < or > → hollow circle (endpoint not included)

• ≤ or ≥ → solid/dark circle (included)

Direction:

• x > a: shade to the right

• x < a: shade to the left

• a < x ≤ b: between a and b, left open, right closed