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Quadratic Equations

If f(x) is a quadratic polynomial [general form ax² + bx + c = 0], then the equation f(x) = 0 is called a quadratic equation. The values of x satisfying f(x) = 0 are called its roots or zeros. These are the values of x which satisfies the equation ax² + bx + c = 0 where a, b and c are real numbers.

 

Solving Quadratic Equations by Factoring

The standard form for a quadratic equation is ax2 + bx + c = 0, where a, b, and c are real numbers and a 0.

For example, the following are examples of quadratic equations:

x2 + 6x + 5 = 0, 3x2 - 2x = 0, and x2+ 4 = 0.

 

Illustration 1 :

Solve x2 – 7x + 12 = 0

Solution

We have to find the factors of ac. In this case ac = 12.
The factors of 12 are 4 and 3.
Hence we can write the equation as: x2 – 4x – 3x + 12 = 0
Taking two terms at a time, we get x(x – 4) – 3(x – 4) = 0 ⇒ (x – 3)(x – 4) = 0.
Putting these separately equal to 0 we get x – 3 = 0, x = 3 and x – 4 = 0, x = 4.
Hence the equation has 2 solutions x = 3 and x = 4.
The solutions of an equation are also called the roots of the equation.

 

Not all quadratic equations can be solved in this way, as ac may not have factors. Also, remember that a quadratic equation has at most two real roots. Sometimes it may have no real root.  For example, the equation x2 + 4 = 0 has no real root; since x2 = - 4 does not have real roots.

 

Solving Equations by using the Quadratic Formula

If a quadratic expression is not easily factored, then its roots can always be found using the quadratic formula: If ax2 + bx + c = 0 (a 0), then the roots are
x =   [- b + âˆš (b2 - 4ac) ]/2a and x =   [ -b -  (b2 - 4ac) ]/2a

                                        

Important points

b2 – 4ac is called the discriminant and is denoted by the symbol or is represented by the letter D. Following are some of the important points relating to the discriminant and its relation with the nature of the roots.

  • If > 0, then both the roots will be real and unequal and the value of roots will be b ±√Δ/2a.
  • If is a perfect square, then roots are rational otherwise they are irrational.
  • If = 0, then roots are real, equal and rational. In this case the value of roots will be – b/2a.
  • If < 0, then roots will be imaginary, unequal and conjugates of each other.
  • If α  and β are the roots of the equation ax2 + bx + c = 0, then sum of the roots i.e. α + β  = -b/a.
  • If α  and β are the roots of the equation ax2 + bx + c = 0, then product of the roots i.e. αβ = c/a.
  • If α  and β , the two roots of a quadratic equation is given, then the equation will be

x2(α + β)x + αβ = 0.            

The equation is x2 – (sum of roots)x + product of roots = 0

  • If in the equation b = 0, then roots are equal in magnitude, but opposite in sign.
  • If a = c, then roots are reciprocal of each other.
  • If c = 0, then one of the roots will be zero.
  • If one root of a quadratic equation be a complex number, the other root must be its conjugate complex number i.e. = j + -k, then = j - -k = j + ik and β = j – ik
Illustration:

If one root of the quadratic equation 8x2 – 28x + z = 0 is six times the other. Find the value of z.

Solution

In this equation a = 8, b = - 28 and c = z.
From the formula, sum of the roots = - (-28)/8 = 7/2
If one root is , the other root is 6 and the sum of roots will be 7.
Hence 7 = 7/2 ⇒ = ½. Other root will be ½ × 6 = 3.
Product of the roots = c/a = z/8.
Product of roots α.6α = 6α2. 6(1/2)2  hence 6/4 = z/8.
z = 12.

 

Illustration:

If 16x2 – 24x + m = 0 have equal roots, find the value of m.

Solution

Because it has equal roots, discriminant b2 – 4ac = 0.
Hence, (-24)2 – 4.16.m = 0 ⇒ 576 = 64m.
64m = 576 ⇒ m = 576/64 = 9.
 

 

Example

If p and q are the roots of the equation x2 + px + q = 0, then which of the following is true?
        1. p = 1                            2. p = 1 or 0 or –
        3. p = –2                          4. p = –2 or 0

Solution

Since p and q are the roots of the equation x2 + px + q = 0,
        Then p2 + p2 + q = 0        and   q2 + pq + q = 0
        →     2p2 + q = 0      and   q(q + p + 1) = 0
        →     2p2 + q = 0      and   q = 0 or q = –  p – 1         
        when we take 2p2 + q = 0 and q = 0, we get p = 0.
When we take 2p2 + q = 0 and q = –  p – 1, we get 2p2 – p – 1 = 0, which gives us p =1 or p = - ½.
Hence there can be three values for P i.e. p = 0 or 1 or - ½. Thus 3rd option is our answer. 

    





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