Question

Without solving determine the number of real solutions for each quadratic equation

b^2-4b+3=0
2n^2 + 7 = -4n + 5
x - 3x^2 = 5+ 2x - x^2

Answer 1

b^2-4b+3=0

b²-3x-b+3=0

b(b-3)-1(b-3)=0

(b-3)(b-1)=0

either

b=3 or b=1

.

2n^2 + 7 = -4n + 5

2n²+4n+7-5=0

2n²+4n+2=0

2(n²+2n+1)=0

(n+1)²=0/2

:.n=-1

.

x - 3x^2 = 5+ 2x - x^2

0=5+ 2x - x^2-x +3x^2

0=5+x+2x²

2x²+x+5=0

comparing above equation with ax²+bx +c we get

a=2

b=1

c=5

x={-b±√(b²-4ac)}/2a ={-1±√(1²-4×2×5)}/2×1

={-1±√-39}/2

Answer 2

• Complete the square to determine the number of solutions

#1

• b² -4b + 3 = 0
• b² - 4b + 4 = 1
• (b - 2)² = 1

• Two solutions

#2

• 2n² + 7 = -4n + 5
• 2n² + 4n + 2 = 0
• n² + 2n + 1 = 0
• (n + 1)² = 0

• One solution

#3

• x - 3x² = 5 + 2x - x²
• 3x² - x² + 2x - x + 5 = 0
• 2x² + x + 5 = 0
• x² + x² + 2*1/2x + 1/4 + 4 3/4 = 0
• x² + (x + 1/2)² + 4 3/4 = 0

• No solutions