Quadratics Double Root

Question

For the equation 2x^2-4x=b,

a) give the value for b that gives the equation a double root. Show your work and explain your reasoning.

b) Give a value for b that gives the equation rational solutions. Show your work and explain your reasoning.

Answer 1

For part a) we want to build a perfect square trinomial to make the double root occur.

2x² - 4x - b = 0

2(x² - 2x - b/2) = 0

We want -b/2 = 1 so make the perfect square trinomial. We would then have 2(x² - 2x + 1) = 0 or 2(x - 1)² = 0, which gives a double root at x = 1. -b/2 = 1 implies that b = -2.

Another way to solve part a) would be to set the discriminant equal to 0.

b² - 4ac = 0

(-4)² - 4(2)(-b) = 0

16+8b = 0

b = -2.

For part b) we want the quadratic to be factorable in order to get rational solutions out of it.

2x² - 4x - b factorable

One example: 2x² - 4x - 6 = 2(x² - 2x - 3) = 2(x - 3)(x + 1) = 0 would give x = 3 and x = -1 as rational solutions. The value of b that I used for this example was b = 6.

Answer 2

A) If a quadratic function is a perfect square with real roots, then it has a double root. Complex roots come in conjugate pairs.

so 2x² - 4x - b = 0

or 2(x² - 2x - b/2 = 0)

since x² - 2x + 1 = 0 we can say that -b/2 =1, so b= -2 would give the function a double root at x= 1 since (x² - 2x - b/2 ) = ( x -1 )² = 0 when x = 1

b) b = -2 gives one rational solution, x = 1. We need rational solutions. With an S. So in our quadratic formula derived from completing the square on ax^2 + bx + c = 0 and solving x, we will see that if sqrt(b^2 -4ac) is a non-zero integer, the function will have two rational roots. In our quadratic formula, a=2, b = -4, and c = -d. Let,s solve for d since we used b in the quadratic formula. So sqrt(16+8d) = N where N is a nonzero integer. Say 16+ 8d = 64 then d = 6 and N =8. And since we renamed b, d. So let’s say b = 6.

checking in 2x² - 4x - b = 0 : 2x² - 4x - 6= 0 => 2(x²- 2x- 3)= 0 => x = 3 or -1 check