Solve the quadratic equation and give your answer correct to 2 decimal places 5x x 2 )= 3

5x(x + 2) = 3
⇒ 5x2 + 10x = 3
⇒ 5x2 + 10x - 3 = 0Here a = 5, b = 10, c = -3

D = b2 - 4ac

= 160

∴ x = `(-b ± sqrt(b^2 - 4ac))/(2a)`

= `(10 ± sqrt(160))/(2 xx 5)`

= `(-10 ± sqrt(16 xx 10))/(10)`

= `(-10 ±4sqrt(10))/(10)`

= `(-10 ± 4(3.162))/(10)`

= `(-10 ± 12.648)/(10)`

∴ x1 = `(-10 + 12.648)/(10)`

= `(2.648)/(10)`
= 0.2648

x2 = `(-10 - 12.648)/(10)`

= `(-22.648)/(10)`= -2.2648

∴ x = 0.26, -2.26.

Page 2

Solve the following equation by using quadratic formula and give your answer correct to 2 decimal places : 4x2 – 5x – 3 = 0

Given equation 4x2 – 5x – 3 = 0
Comparing with ax2 + bx + c = 0, we have
a = 4, b = -5, c = -3

∵ x = `(-b ± sqrt(b^2 - 4ac))/(2a)`

= `(-(-5) ± sqrt((-5)^2 - 4 xx 4 xx (-3)))/(2 xx 4)`

= `(5 ± sqrt(25 + 48))/(8)`

= `(5 ± sqrt(73))/(8)`

= `(5 ± 8.544)/(8)`

= `(5 + 8.544)/(8) or (5 - 8.544)/(8)`

= `(13.544)/(8) or (-3.544)/(8)`= 1.693 or -0.443

= 1.69 or -0.44.  ...(correct to 2 demical places)

Concept: Quadratic Equations

  Is there an error in this question or solution?

Page 3

`2x - (1)/x = 1`

⇒ 2x2 - 1 = 7x
⇒ 2x2 - 7x - 1 = 0   ....(i)
Comparing (i) with ax2 + bx + c, we get,
a = 2, b = -7, c = -1

∵ x = `(-b ± sqrt(b^2 - 4ac))/(2a)`

⇒ x = `(-(-7) ± sqrt((-7)^2 - 4(2) xx (-1)))/(2 xx 2)`

⇒ `(7 ± sqrt(49 + 8))/(4)`

⇒ `(7 ± sqrt(57))/(4)`

⇒ `x = (7 + sqrt(57))/(4) or x = (7 - sqrt(57))/(4)`

⇒ `x = (7 + 7.55)/(4) or x = (7 - 7.55)/(4)`

⇒ `x = (14.55)/(4) or x = (-0.55)/(4)`
⇒ x = 3.64 or x = -0.14.