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1. Determine the discriminant and nature of roots of each quadratic equation.

a.) x2 -6x + 9=0
b.) x2 - 4x +3=0
c.) x2 -7x - 4=0
d.) 2x2 + 3x + 5=0​

User Suede
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1 Answer

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Answer:

Please check the explanation.

Explanation:

The general quadratic equation is

ax²+bx+c=0

The discriminant = D = b² - 4ac

When b² - 4ac = 0 there is one real root.

When b² - 4ac > 0 there are two real roots.

When b² - 4ac < 0 there are two complex roots.

1) x² -6x + 9=0

On comparing with given quadratic equation x² -6x + 9=0

a = 1, b=-6, c=9

D = b² - 4ac

= (-6)² - 4(1)(9)

= 36 - 36

= 0

D = 0

Thus, there is one real root of quadratic equation x² -6x + 9=0.

2) x² -4x + 3=0

On comparing with given quadratic equation x² -4x + 3=0

a = 1, b=-4, c=3

D = b² - 4ac

= (-4)² - 4(1)(3)

= 16 - 12

= 4

D > 0

Thus, there are two real roots of quadratic equation x² -4x + 3=0.

3) x² -7x - 4=0

On comparing with given quadratic equation x² -7x - 4=0

a = 1, b=-7, c=-4

D = b² - 4ac

= (-7)² - 4(1)(-4)

= 49 + 16

= 65

D > 0

Thus, there are two real roots of quadratic equation x² -7x - 4=0

4) 2x² +3x +5=0

On comparing with given quadratic equation 2x² +3x +5=0

a = 2, b=3, c=5

D = b² - 4ac

= (3)² - 4(2)(5)

= 9 - 40

= -31

D < 0

Thus, there are two complex roots of quadratic equation 2x² +3x +5=0

User Pushp Vashisht
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