Question

If X=3,X=4 are roots of ax²-7x+b=0 then find the value of a and b

Answer 1

Given

$\sf x=3\: and \:x=4\: are\: the \:roots\\ \sf of\: p(x)=ax^2 -7x+b=0$

So by factor theorem ,

$\sf p(3)=0 \\ \sf</p><p>p(4)=0$

substituting x=3 in the equation

$a(3×3)-7(3)+b=0$

$9a+b-21=0$

$9a+b=21...........(1)$

Substituting x=4

$a(4×4)-7(4)+b=0$

$16a+b-28=0$

$16a+b=28.............(2)$

(2)-(1)

$16a+b-9a-b=28-21$

$7a=7$

$\huge \boxed{a=1}$

substituting a =1 in (2)

$16+b=28$

$\huge\boxed{b=12}$

Therefore values of a and b are 1,12

Answer 2

Answer:

a = 1 , b= 12

Step-by-step explanation:

if x = 3

ax²-7x+b=0

a(3²) - 7 (3) + b = 0

9a +b = 21 >>>>>>> 1

if x = 4

ax² - 7x + b = 0

16a + b = 28 >>>>>> 2

by 2 - 1

16a - 9a + b - b = 28 -21

7a = 7

a =1

in eq1

b = 21 - 9a

b = 21 -9

b = 12

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