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Lisa [10]
3 years ago
15

If x=2/3 and x=-3 are the roots of the equation ax^2+7x+b=0 find the values of a and b

Mathematics
1 answer:
Gnesinka [82]3 years ago
7 0

<u>ANSWER:  </u>

If x = \frac{2}{3} and x = -3 are the roots of the equation, then values of a and b are 3, -6

<u>SOLUTION: </u>

Given, quadratic equation is a x^{2}+7 x+b=0 and its roots are -3 , \frac{2}{3}

We know that, for any quadratic equation of form a x^{2}+b x+c=0 with roots x_{1} and x_{2} then,

Sum of roots (x_{1} +x_{2}) = \frac{-b}{a}

Product of roots ( x_{1} \times x_{2}) = \frac{c}{a}

Now, for given quadratic equation x_{1} = -3 and x_{2} = \frac{2}{3}

hence Sum of roots = \frac{-7}{a}

x_{1} + x_{2} = \frac{-7}{a}

-3 + \frac{2}{3} = \frac{-7}{a}

on solving we get "a" = 3

Now product of roots = x_{1} \times x_{2} = \frac{b}{3}

-3 \times \frac{2}{3} = \frac{b}{3}

b = -6

hence, the values of a and b are 3, -6.

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zepelin [54]
The complete question is
<span>-2 2/3, -5 1/3, -10 2/3, -21 1/3, -42 2/3, ... Which formula can be used to describe the sequence?

a. f(x + 1) = –2f(x)
b. f(x + 1) = f(x)
c. f(x + 1) = -f(x)
d. f(x + 1) = 2f(x)

case a) </span>f(x + 1) = –2f(x)
for f(x)=-2 2/3
f(x+1)=-2*[-2 2/3]=5 1/3
5 1/3 is not -5 1/3
therefore
this is not the formula

case b) f(x + 1) = f(x)
for f(x)=-2 2/3
f(x+1)=-2 2/3
-2 2/3 is not -5 1/3
therefore
this is not the formula

case c) f(x + 1) = -f(x)
for f(x)=-2 2/3
f(x+1)=2 2/3
2 2/3 is not -5 1/3
therefore
this is not the formula

case d) f(x + 1) =2f(x)
for f(x)=-2 2/3
f(x+1)=2*[-2 2/3]=-5 1/3

for f(x)=-5 1/3
f(x+1)=2*[-5 1/3]=-10 2/3

for f(x)=-10 2/3
f(x+1)=2*[-10 2/3]=-21 1/3

for f(x)=-21 1/3
f(x+1)=2*[-21 1/3]=-42 2/3

therefore 
the answer is the option
d. f(x + 1) = 2f(x)

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