Question

Find the value of a and b in quadratic equation ax^2 -13x +b =0, where x=3÷2 and x=2÷ 3 are the roots
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Answer 1

The roots of given equation I. e. the solutions of the equation are given.
For x=3/2
$a {x}^{2} - 13x + b = 0 \\ put \: x = \frac{3}{2} \\ a \times ( { \frac{3}{2}) }^{2} - 13 \times \frac{3}{2} + b = 0 \\ \frac{9}{4} a - \frac{39}{2} + b = 0 \\ multiplying \: by \: 8on \: both \: sides \: \\ 18a - 156 + b = 0 \\ 18a + b = 156...................(1)$
Now for x=2/3
$put \: x = \frac{2}{3} \\ a {x}^{2} - 13x + b = 0 \\ a \times ( { \frac{2}{3}) }^{2} - 13 \times \frac{2}{3} + b = 0 \\ \frac{4}{9} a - \frac{26}{3} \ + b = 0 \\ multiplying \: by \: 27on \: both \: sides \: \\ 12a - 234 + b = 0 \\ 12a + b = 234...............(2)$
Now, equation (1)-(2),we get
18a+b=156
-12a+b=234
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6a. = -78
therefore.' a= -13'
and. ' b=390'