Q11 of 168 Page 7

For what values of a and b, and x = — 2 are solutions of the equation ax2 + bx — 6 = 0.

Put x = 3/4


ax2 + bx — 6 = 0





9a + 12b – 96 = 0 divide by 3


3a + 4b – 32 = 0


3a + 4b = 32 (1)


Put x = –2


ax2 + bx — 6 = 0


a(–2)2 + b(–2) – 6 = 0


4a – 2b – 6 = 0


4a – 2b = 6 (2)


Eliminate (1) and (2)


3a + 4b = 32


4a – 2b = 6 ×2



a = 4


Put a = 4 in equation (1).


3a + 4b = 32


3(4) + 4b = 32


12 + 4b = 32


4b = 32 – 12


4b = 20


b = 5


More from this chapter

All 168 →
9

For what value of k, is the solution of the equation

kx2 – x – 2 = 0.

 10

For what value of k, is a solution of the equation 3x2 + 2kx — 3 = 0

 12

For what value of k, x = a is a solution of the equation

x2 – (a + b) x + k = 0.

 13

Determine the value of k, a and b in each of the following quadratic equation, for which the given value of x is the root of the given quadratic equation:

(i) kx2 — 5x + 6 = 0 ; x = 2


(ii) 6x2 + kx — √6 = 0;


(iii) ax2 — 13x + b = 0; x =2 and x = —2 find a, b


(iv) ax2+ bx – 10 = 0; and