Question

If x = 5 and x = 4 are the roots of the equation ax^2+ bx + 20 = 0, find the values of a and b.​

Answer 1

Step-by-step explanation:

Given :-

x = 5 and x = 4 are the roots of the equation ax^2+ bx + 20 = 0

To find :-

Find the values of a and b ?

Solution :-

Given equation is ax²+bx+20 = 0

Given roots = 5 and 4

If x = 5 is a root then it satisfies the given equation

=> a(5)²+b(5)+20 = 0

=> 25a+5b+20 = 0

=> 5(5a+b+4) = 0

=> 5a+b+4 = 0/5

=> 5a +b+4 = 0

=> 5a+b = -4 -------------------(1)

and

If x =4 is a root then it satisfies the given equation

=> a(4)²+b(4)+20 = 0

=> 16a+4b+20 = 0

=> 4(4a+b+5) = 0

=> 4a+b+5 = 0/4

=> 4a +b+5 = 0

=> 4a + b = -5 -------------------(2)

On Subtracting (2) from (1)

5a + b = -4

4a + b = -5

(-) (-) (+)

_________

a + 0 = 1

_________

=> a = 1

On Substituting the value of a in (1) then

=> 5(1)+b = -4

=> 5+b = -4

=> b = -4-5

=> b = -9

We have,

a = 1 and b = -9

Answer:-

The values of a and b are 1 and -9 respectively.

Check:-

If a = 1 and b = -9 then the equation becomes

(1)x²+(-9)x+20 = 0

=> x²-9x+20 = 0

=> x²-5x-4x+20 = 0

=> x(x-5)-4(x-5) = 0

=> (x-5)(x-4) = 0

=> x-5 = 0 or x-4 = 0

=> x = 5 or x = 4

Therefore, x = 5 and x = 4

Verified the given relations in the given problem

Used Concept:-

If a number is a root then it satisfies the given equation (.i.e. LHS = RHS )