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Question Number 136667 by mathlove last updated on 24/Mar/21

i f f (x) = {a x}^{2} + b x + c a n d f (5) = - 3 f (2)

a n i n t e r s i c t i o n p o i n t (- 4, 0)

a n t h e X - a x i s f a i n d a n o t h e r p o i n t

t h e X - a x i s

Answered by MJS_new last updated on 25/Mar/21

I should have finished it

(1) 25 a + 5 b + c = - 3 (4 a + 2 b + c) \Leftrightarrow 37 a + 11 b + 4 c = 0

(2) 16 a - 4 b + c = 0

(1) \Rightarrow c = - \frac{37 a + 11 b}{4}

\Rightarrow (2) \frac{27}{4} a - \frac{27}{4} b = 0 \Rightarrow b = a \Rightarrow c = - 12 a

\Rightarrow f (x) = a (x^{2} + x - 12) = a (x - 3) (x + 4)

\Rightarrow zeros are - 4 and 3

Answered by ajfour last updated on 25/Mar/21

25 a + 5 b + c = - 3 (4 a + 2 b + c)

\Rightarrow 37 a + 11 b + 4 c = 0

\frac{c}{a} = - \frac{11}{4} (\frac{b}{a}) - \frac{37}{4}

x = - \frac{b}{2 a} \pm \frac{\sqrt{b^{2} - 4 a c}}{2 a}

x = - \frac{b}{2 a} \pm \sqrt{\frac{1}{4} {(\frac{b}{a})}^{2} - \frac{c}{a}}

x = \frac{b}{a} {- \frac{1}{2} \pm \sqrt{\frac{1}{4} + \frac{11}{4 (b / a)} + \frac{37}{4 {(b / a)}^{2}}}}

s a y \frac{b}{a} = p

{(\frac{- 4}{p} + \frac{1}{2})}^{2} = \frac{1}{4} + \frac{11}{4 p} + \frac{37}{4 p^{2}}

\frac{16}{p^{2}} - \frac{4}{p} = \frac{11}{4 p} + \frac{37}{4 p^{2}}

\frac{27}{4 p^{2}} - \frac{27}{4 p} = 0

\Rightarrow p = 1

x = {- \frac{1}{2} \pm \sqrt{\frac{1}{4} + \frac{11}{4} + \frac{37}{4}}}

x = - \frac{1}{2} \pm \frac{7}{2}

x = - 4 o r 3

o t h e r p o i n t o n x - a x i s i s

(3, 0)

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