Question

If x $\frac{2}{3}$ and x= - 3 are roots of the quadratic equation ax² + 7x+ b= 0, find the values of a and b.

Answer 1

$a {x}^{2} + 7x + b = 0$


x=$\frac{2}{3}$

$a \times ( \frac{2}{3} ) {}^{2} + 7 \times \frac{2}{3} \times + b = 0$


$a \times \frac{4}{9} + \frac{14}{3} + b = 0$


$\frac{4a}{9} + \frac{14}{3} + b = 0$
---------(1)

x=-3 .

a×(-3)^2 + 7×(-3)+b=0

9a -21 +b =0 -------(2)

Equalising equation 1 and 2 .

$\frac{4a}{9} + \frac{14}{3} + b = 9a - 21 + b$


$\frac{4a}{9} - 9a + b - b \: = - 21 - \frac{14}{3}$


$\frac{4a - 81a}{9} = \frac{ - 63 - 14}{3}$


$\frac{ - 77a}{9} = \frac{ - 77}{3}$



$\frac{a}{9} = \frac{1}{3}$



3a =9

a=9/3

a= 3

Using this value of a in equation 2 .

9a-21+b =0

9×3-21 + b=0

27-21+b=0

6+b=0

b=0-6

b=-6

$\bf{Hope\:it\:helps}$

Answer 2

Solution :

Given quadratic equation :

ax² + 7x + b = 0 ---( 1 )

i ) Put x = 2/3 in equation ( 1 ), we get

a(2/3)² + 7(2/3) + b = 0

=> 4a/9 + 14/3 + b = 0

Multiply each term by 9 , we get

=> 4a + 42 + 9b = 0

=> 4a + 9b = -42 ---( 2 )

ii ) Put x = -3 in equation ( 1 ) , we get

a(-3)² + 7( -3 ) + b = 0

=> 9a - 21 + b = 0

=> b = 21 - 9a ---( 3 )

iii ) Now , substitute b = 21 - 9a in

equation ( 2 ) , we get

4a+ 9( 21 - 9a ) = -42

=> 4a - 189 - 81a = -42

=> -77a = -42 - 189

=> -77a = -231

=> a = ( -231 )/( -77 )

=> a = 3

Put a = 3 in equation ( 3 ), we get

b = 21 - 9×3

=> b = 21 - 27

=> b = -6

Therefore ,

a = 3 , b = -6

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