Question

If one root of the equation x^2-3x+m=0 exceeds the other by 5 then the value of m is equal to

Answer 1

m=-4

Step-by-step explanation:

You can see the answer in the picture shown above as it is a well illustrated one

Answer 2

Given,

One root of the equation x^2-3x+m=0 exceeds the other by 5.

To find,

The value of m

Solution,

Let the one root be m and then the other root will be m+5.

We know that if ax²+bx+c=0 is the quadratic equation then the sum of roots is represented by  α+β=-b/a and the product of roots is represented by  αβ=c/a

Similarly, in the equation, x²-3x+m=0 sum of roots would be -

⇒   t + t+5 = -(-3)

⇒  2t+5=3   2t+2=0   Equation 1

⇒  2t+2=0

⇒  2t=-2

⇒  t=-1

So if the one root t = -1 then the other root would be t+5 i.e. -1+5 = 4

In the equation, x²-3x+m=0 product of roots would be ⇒ m

And the Product of roots -1 and 4 would be -4.

Therefore, If one root of the equation x^2-3x+m=0 exceeds the other by 5 then the value of m is equal to -4.