Question

Given the difference between the roots of the quadratic equation `x^2`+ `6x` + h - 3 = 0 is 4, where h is a constant. Find the value of h?

Answer 1

`h = 8`

Explanation:

Given: `x^2+6x+h-3`

The given equation is in standard form where `a = 1, b = 6 and c = h-3`

We are given two roots; let them be `r_1 and r_2` and we are given `r_2 = r_1+4`.

We know that the axis of symmetry is:

`s = -b/(2a)`

`s = -6/(2(1))`

`s = -3`

The roots are symmetrically placed about the axis of symmetry, which means that the first root is axis of symmetry minus 2 and the second root is the axis of symmetry plus 2:

`r_1 = -3-2 = -5` and `r_2 = -3+2 = -1`

Therefore, the factors are:

`(x+5)(x+1) = x^2+6x+5`

We can write the following equation to find the value of h:

`5 = h - 3`

`h = 8`

Answer 2

Another method

Explanation:

We have 2 roots `r_1,r_1+4`. So multiply them and compare coefficients

`(x+r_1)(x+r_1+4) = x^2+6x+(h-3)`
`x^2+(2r_1+4)x+r_1(r_1+4) = x^2+6x+(h-3)`
`2r_1+4 = 6`
`r_1 = 1`
`1(1+4) = h-3`
`h = 8`

Answer 3

`h=8`

Explanation:

we have

`x^2+6x+h-3=0`

the difference in roots is 4

so if one root is `alpha`

the other is `alpha+4`

now for any quadratic

`ax^2+bx+c=0`

with roots

`alpha, beta`

`alpha+b=-b/a`

`alphabeta=c/a`

so;

`alpha+alpha+4=-6`

`2alpha=-10=>alpha=-5`
hence

`beta=alpha+4=-1`

`alphabeta=-5xx-1=h-3`

`:.h-3=5`

`=>h=8`