a) find, as surds, the roots of the equation

2(x+1)(x-4)-(x-2)^2 = 0

b) use algebra to solve

(x-1)(x+2)=18

c) find the values of k for whichkx^2 +8x + k = 0

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- Oct 8th 2005, 08:41 AMx-disturbed-xmixed quadratics
a) find, as surds, the roots of the equation

**2(x+1)(x-4)-(x-2)^2 = 0**

b) use algebra to solve

**(x-1)(x+2)=18**

c) find the values of k for which**kx^2 +8x + k = 0** - Oct 8th 2005, 08:59 AMctelady
2(x+1)(x-4)-(x-2)^2 = 0 FOIL -->

2[x^2 -3x -4] -[x^2 -4x +4] = 0

2x^2 -6x -8 -x^2 +4x -4 = 0

2x^2 -x^2 -6x +4x -8 -4 = 0

x^2 -2x -12 = 0 Quadratic equation from here with a = 1, b = -2, c = -12

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(x-1)(x+2)=18 FOIL -->

x^2 +x -2 = 18 Set equal to zero [standard form for quadratics] -->

x^2 +x -20 = 0 Factor -->

(x+5)(x-4) = 0 Set each factor equal to zero and solve. - Oct 16th 2005, 08:25 PMedcity
AS for part C of your qns, you are looking at the equation :

**kx^2 +8x + k = 0**This will require some of your basics in quadratic equations. For a simple quadratic equation, ax^2 + bx + c = 0, this will be true if b^2 - 4ac term is greater or equal to zero. So the solution to this problem follows the same concept.

kx^2 +8x + k = 0

Thus for this equation to be true, first:

(8^2) - 4(k)(k) > 0

64 - 4k^2 > 0

16 - k^2 > 0

k^2 - 16 > 0

(k - 4) (k + 4) > 0

Thus, by refering to the number line:**k > 4**and**K < -4**

This will be the suggested solution.