Julie is solving the equation x2 + 5x + 6 = 0 and notices that the discriminant b2 - 4ac has a value of 1. This tells her that the equation has?this is a one variable equation

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hala718 | High School Teacher | (Level 1) Educator Emeritus

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If b^2 - 4ac is a positive number, then that means that the equation has two real roots ( solutions).

Let us verify.

x^2 + 5x + 6 = 0

==> a = 1, b = 5, c = 6

==> b^2 - 4ac = 25 - 4*1*6 = 1.

Let us determine the roots.

==> x1= (-b + sqrt(b^2 -4ac)/ 2a

==> x1= (-5 + 1)/2 = -4/2 = -2

==> x2= (-5-1)/2 = -6/2 = -3.

Then we notice that the function has 2 real roots.

On the other hand.

If b^2 - 4ac = 0 ==> Then, the function has only one root.

If b^2 - 4av < 0 ==> Then, the function has two complex roots.

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neela | High School Teacher | (Level 3) Valedictorian

Posted on

To solve the equation x^2+5c+6 = 0.The discriminant is 1.

Solution: A quadratic equation  2nd degree of a single variable x is like ax^+bx+c = 0, where a, b and c are coefficients.The discriminant of this equation is b^2-4ac. And the equation has two roots x1 and x2 given by: x1 = {-b+sqrt(b^2-4ac})}/2a  and x2 =  {-b-sqrt(b^2-4ac})}/2a. Or x1, x2 = (-b+or-sqrt(discriminant)}/2a.

In our case, x^2+5c+6 = 0, or1*x^25x+6 = 0, a = 1, b= 5 and c= 5. So the discriminant of x^2+5x+6 = 0 is 5^2-4*1*6 = 25-24 = 1.

Therefore the equation has two distinct roots  given by:

 x1 = {-5+sqrt(discriminant 1)}/2*1} =  {-5+1}/2 = -2

x2 = {-5-sqrt(discriminant 1)}/2*1} =  {-5-1}/2 = -3.

So the roots are of the equation has two roots.

Again be reminded that discriminant 1 does not mean the equation has single variable. The discriminant is a different concept. The discriminant gives a measure of (x1-x2)^2 , the square of the differences of the roots.

Hope this helps.

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atyourservice | Student, Grade 11 | (Level 3) Valedictorian

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the fact that the discriminant is 1 tells her that the problem has 2 real solutions as the number is bigger than 0, when the number is 0 it has 1 real solution and when the number is negative it means the problem has 0 real solutions

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