Form 4 Add Math Chapter 2
2.2 Solving Quadratic Equation
A quadratic equation can be solved by using: 1. Factorization i) Arrange the unknowns in general form (ax2+bx+c=0) and factorize it 2. Completing the square i) x2-6x+4=0 ii) x2-6x=-4 iii) x2-6x+-622=-4+-622 iv) x2-6x+-32=-4+-32 v) x-32=5 vi) x-3=±5 vii) x=3±5 viii) x=3+5 OR 3-5 ix) x=5.236/0.7369
3. Quadratic formula x=-b±b2-4ac2a * Forming quadratic equation from given roots
SOR=sum of roots POR=product of roots
SOR=-ba POR=ca
.x2-α+βx+αβ=0
.x2-SORx+POR=0 * Types of roots b2-4ac>0 | Two different roots | Straight line intersects a curve at two different points | b2-4ac=0 | Two equal roots | Straight line touches the curve at one point/ tangent to the curve | b2-4ac<0 | No real roots | Straight line does not intersect the curve |
Example Questions:
1. Show that the quadratic equation x2=21-kx-9-k2 has two real roots for k≤-4.
Solution:
.x2-21-kx+9+k2=0
.a=1, b=-2+2k, c=9+k2
Since it has 2 real roots,
.b2-4ac≥0 {It didn’t mention that the 2 roots are equal roots or different roots, so it can be >0/=0 that becomes ≥0}
.(-2+2k)2-4(1)(9+k2)≥0
.4-8k+4k2-36-4k2≥0
.-8k-32≥0
.-8k≥32
.-k≥4
.k≤-4
A quadratic equation can be solved by using: 1. Factorization i) Arrange the unknowns in general form (ax2+bx+c=0) and factorize it 2. Completing the square i) x2-6x+4=0 ii) x2-6x=-4 iii) x2-6x+-622=-4+-622 iv) x2-6x+-32=-4+-32 v) x-32=5 vi) x-3=±5 vii) x=3±5 viii) x=3+5 OR 3-5 ix) x=5.236/0.7369
3. Quadratic formula x=-b±b2-4ac2a * Forming quadratic equation from given roots
SOR=sum of roots POR=product of roots
SOR=-ba POR=ca
.x2-α+βx+αβ=0
.x2-SORx+POR=0 * Types of roots b2-4ac>0 | Two different roots | Straight line intersects a curve at two different points | b2-4ac=0 | Two equal roots | Straight line touches the curve at one point/ tangent to the curve | b2-4ac<0 | No real roots | Straight line does not intersect the curve |
Example Questions:
1. Show that the quadratic equation x2=21-kx-9-k2 has two real roots for k≤-4.
Solution:
.x2-21-kx+9+k2=0
.a=1, b=-2+2k, c=9+k2
Since it has 2 real roots,
.b2-4ac≥0 {It didn’t mention that the 2 roots are equal roots or different roots, so it can be >0/=0 that becomes ≥0}
.(-2+2k)2-4(1)(9+k2)≥0
.4-8k+4k2-36-4k2≥0
.-8k-32≥0
.-8k≥32
.-k≥4
.k≤-4