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Quadratic Challenging Questions 3/3

Find the range of values of m for which the line x+2y=m meets the curve x(x+y)+8=0. State the values of m for which the line is a tangent to the curve.

x(x+y)+8=0 –> y=-8/x-x –equation 1

2y=m-x –> y=(m-x)/2 –equation 2

equation 1 = equation 2

since the 2 lines meet –> equation 1 = equation 2

-8/x-x = (m-x)/2

-16-2x2=mx- x2

0= x2+mx+16

Look out for patterns.. if possible, always arrange equation in

(curve) ax2+bx+c or

(straight line) mx+c or

(cubic curve) (dx+e)(ax2+bx+c)

b2-4ac

= m2-4(16)

= (m-8)(m+8)

a) b2-4ac
= (m-8)(m+8) ≥ 0 –> m ≤ -8 or m ≥ 8 for equation 1 to meet equation 2 🙂

(since equation 1 and equation 2 meet at 1 or 2 or more points,
b2-4ac ≥ 0)

b) b2-4ac
= (m-8)(m+8) = 0 –> m=8 or -8 for line to meet curve at only 1 point 🙂

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