Answer to Question #233063 in Algebra for dizzo

Question

28. (a) Given the equation

x^2+qx+2a^2-b^2+ab=0

i) Factorize the independent term of the equation /3mrks

ii) After determining the product of roots of (E), deduce the roots given that one of them is

a +b /3mrks

iii) Deduce q in terms of a and b. /2mrks

(b) Determine two numbers whose S = 4a -b and P 3a^2- 2b^2- 5ab /4mrks

(c) Given the equation 3x^2+ kx + 8 = 0. Determine the roots of this equation if their ratio is 1.5.

Deduce the value of k.

Answer 1

$x^2+qx+2a^2-b^2+ab=0$

i)

$2a^2-b^2+ab=a^2-b^2+a^2+ab$

$(a-b)(a+b)+a(a+b)=(2a-b)(a+b)$

ii)

$x_1x_2=\dfrac{(2a-b)(a+b)}{1}=(2a-b)(a+b)$

If $x_1=a+b,$ then $x_2=2a-b.$

$\{ a+b, 2a-b\}$

iii)

$x_1+x_2=-\dfrac{q}{1}=-q$

$q=-(a+b+2a-b)$

$q=-3a$

(b)

$x_1+x_2=4a-b$

$x_1x_2=3a^2-2b^2-5ab$

$x^2-(4a-b)x+3a^2-2b^2-5ab=0$

$D=(4a-b)^2-4(3a^2-2b^2-5ab)$

$=16a^2-8ab+b^2-12a^2+8b^2+20ab$

$=4a^2+12ab+9b^2=(2a+3b)^2$

$x=\dfrac{-(-(4a-b))\pm\sqrt{(2a+3b)^2}}{2}$

$=\dfrac{4a-b\pm(2a+3b)}{2}$

$x_1=\dfrac{4a-b-(2a+3b)}{2}=\dfrac{4a-b-2a-3b}{2}=a-2b$

$x_2=\dfrac{4a-b+(2a+3b)}{2}=\dfrac{4a-b+2a+3b}{2}=3a+b$

(c)

$3x^2+ kx + 8 = 0$

$x_1=1.5x_2$

$x_1x_2=\dfrac{8}{3}$

$1.5x_2^2=\dfrac{8}{3}$

$x_2^2=\dfrac{16}{9}$

$x_2=-\dfrac{4}{3}, x_1=-2, k=x_1+x_2=-\dfrac{10}{3}$

Or

$x_2=\dfrac{4}{3}, x_1=2, k=x_1+x_2=\dfrac{10}{3}$