If ax2 + bx + c = a(x - p)2, then the relation among a, b, c would be?
A. abc = 1
B. b2 = ac
C. b2 = 4ac
D. 2b = a + c
Answer: Option C
Solution(By Examveda Team)
$$\eqalign{ & a{x^2} + bx + c = a{\left( {x - p} \right)^2} \cr & \Rightarrow a{x^2} + bx + c = a{\left( {{x^2} + {p^2} - 2px} \right)^2} \cr & \Rightarrow a{x^2} + bx + c = a{x^2} + a{p^2} - 2apx \cr & {\text{Comparing confficients of }}{x^2}{\text{and }}x \cr & \Rightarrow b = - 2ap \cr & \Rightarrow p = - \frac{b}{{2a}}\,.......(1) \cr & and{\text{ }}c = a{p^2} \cr & \Rightarrow c = a \times \frac{{{b^2}}}{{4{a^2}}}\left[ {{\text{From (i)}}} \right] \cr & \Rightarrow 4ac = {b^2} \cr} $$Related Questions on Algebra
If $$p \times q = p + q + \frac{p}{q}{\text{,}}$$ then the value of 8 × 2 is?
A. 6
B. 10
C. 14
D. 16
A. $$1 + \frac{1}{{x + 4}}$$
B. x + 4
C. $$\frac{1}{x}$$
D. $$\frac{{x + 4}}{x}$$
A. $$\frac{{20}}{{27}}$$
B. $$\frac{{27}}{{20}}$$
C. $$\frac{6}{8}$$
D. $$\frac{8}{6}$$
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