Revision: Quadratic Equations Algebra Maths 1 SSC (English Medium) 10th Standard Maharashtra State Board

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

The set of elements representing the roots of a quadratic equation is called its solution set.

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 

If a quadratic equation contains only two terms where one is a square term and the other is the first power term of the unknown, it is called adjected quadratic equation.

For example :

(i) 4x² + 5x = 0

(ii) 7x² − 3x = 0, etc. 

If the quadratic equation contains only the square of the unknown, it is called a pure quadratic equation.

For example :

(i) x² = 4 

(ii) 3x² − 8 = 0, etc.

For the quadratic equation ax² + bx + c = 0, a ≠ 0; the expression b² − 4ac is called the discriminant and is, in general, denoted by the letter 'D'.

Thus, discriminant D = b² − 4ac.

In the polynomial `f(x)=a_nx^n+a_(n-1)x^(n-1)+...+a_1x+a_0`, 

`a_nx^n,a_(n-1)x^(n-1),...,a_1x`, and `a_0` are known as the terms of the polynomial and `a_n,a_(n-1),...,a_1` and `a_0` are their real coefficients.

For example, `p(x)=3x-2` is a polynomial and 3 is a real coefficient.

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

For a quadratic polynomial

ax² + bx + c (a≠0)

If its zeroes are α and β, then:

\[\alpha+\beta=-\frac{b}{a}\]

\[\alpha\beta=\frac{c}{a}\]

For a cubic polynomial

ax³ + bx² + cx + d, 

\[\alpha+\beta+\gamma=-\frac{b}{a}\]

\[\alpha\beta+\beta\gamma+\gamma\alpha=\frac{c}{a}\]

\[\alpha\beta\gamma=-\frac{d}{a}\]

The quadratic equation whose roots are α and β is

x² − (α+β)x + αβ = 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.

Since α and β are the zeros of the quadratic polynomial f(x) = x² – px + q

`alpha+beta=-"coefficient of x"/("coefficient of "x^2)`

`=(-(-p))/1`

= p

`alphabeta="constant term"/"coefficient of "x^2`

`=q/1`

= q

We have,

`alpha^2/beta^2+beta^2/alpha^2=(alpha^2xxalpha^2)/(beta^2xxalpha^2)+(beta^2xxbeta^2)/(alpha^2xxbeta^2)`

`alpha^2/beta^2+beta^2/alpha^2=alpha^4/(beta^2alpha^2)+beta^4/(alpha^2beta^2)`

`alpha^2/beta^2+beta^2/alpha^2=(alpha^4+beta^4)/(alpha^2beta^2)`

`alpha^2/beta^2+beta^2/alpha^2=((alpha^2+beta^2)-2alpha^2beta^2)/(alpha^2beta^2)`

`alpha^2/beta^2+beta^2/alpha^2=([(alpha+beta)^2-2alphabeta]^2-2(alphabeta)^2)/(alphabeta)^2`

`alpha^2/beta^2+beta^2/alpha^2=([(p)^2-2q]^2-2(q)^2)/q^2`

`alpha^2/beta^2+beta^2/alpha^2=([p^2-2q]^2-2q^2)/q^2`

`alpha^2/beta^2+beta^2/alpha^2=([p^2xxp^2-2xxp^2xx2q+2qxx2q]-2q^2)/q^2`

`alpha^2/beta^2+beta^2/alpha^2=([p^4-4p^2q+4q^2]-2q^2)/q^2`

`alpha^2/beta^2+beta^2/alpha^2=(p^4-4p^2q+4q^2-2q^2)/q^2`

`alpha^2/beta^2+beta^2/alpha^2=(p^4-4p^2q+2q^2)/q^2`

`alpha^2/beta^2+beta^2/alpha^2=p^4/q^2-(4p^2q)/q^2+(2q^2)/q^2`

`alpha^2/beta^2+beta^2/alpha^2=p^4/q^2-(4p^2)/q+2`

Hence, it is proved that `alpha^2/beta^2+beta^2/alpha^2" is equal to "p^4/q^2-(4p^2)/q+2`.

Let a - d, a and a + d be the zeros of the polynomial f(x). Then,

Sum of the zeroes `=("coefficient of "x^2)/("coefficient of "x^3)`

`a-d+a+a+d=(-3b)/a`

`3a=(-3b)/a`

`a=(-3b)/axx1/3`

`a=(-b)/a`

Since a is a zero of the polynomial f(x).

Therefore,

f(x) = ax³ + 3bx² + 3cx + d

f(a) = 0

f(a) = aa³ + 3ba² + 3ca + d

aa³ + 3ba² + 3ca + d = 0

`a((-b)/a)^3+3bxx((-b)/a)^2+3cxx((-b)/a)+d=0`

`axx(-b)/axx(-b)/axx(-b)/a+3xxbxx(-b)/axx(-b)/a+3xxcxx(-b)/a+d=0`

`(-b^3)/a^2+(3b^3)/a^2-(3cb)/a+d=0`

`(-b^3+3b^3-3abc+a^2d)/a^2=0`

2b³ − 3abc + a²d = 0xa²

2b³ − 3abc + a²d = 0

Hence, it is proved that 2b³ − 3abc + a²d = 0.

In the factorisation method, the quadratic expression is written as a product of two linear factors

1. Clear fractions and brackets, if any.

2. Transpose all terms to one side to get the standard form
   ax² + bx + c = 0
3. Factorise the quadratic expression into two linear factors.

4. Put each factor equal to zero (using the zero product rule).

5. Solve the resulting linear equations to obtain the roots.

The completing the square method is used when a quadratic equation cannot be factorised easily.

1. First, write the equation in standard form.
   $ax2+bx+c=0$

2. If a ≠ 1, divide the entire equation by a.

3. Add and subtract \[\left(\frac{b}{2}\right)^2\] to complete the square.

4. Convert the left-hand side into a perfect square.

5. Solve the resulting equation to find the roots.

1. Write the given equation in the standard form

   ax² + bx + c = 0

2. Identify the values of a, b, and c.

3. Find the value of the discriminant

   D = b² − 4ac

4. Substitute the values of a, b, and D in the formula

   \[x=\frac{-b-\sqrt{b^2-4ac}}{2a}\]​

5. Simplify to obtain the roots.

D = b² – 4ac