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# How many roots do the following equations have? -12x^{2 }- 25x + 5 + x^{3 }= 0

**Solution:**

Let f(x) = -12x^{2 }- 25x + 5 + x^{3 }= 0

To find the number of roots for the given equation we need to follow few steps.

**Step 1**: Arrange the given equation in decreasing orders of their powers.

⇒ x^{3}- 12x^{2} - 25x + 5 = 0

**Step 2**: Maximum power of the given equation decides the total number of roots.

⇒ The above polynomial is of the third degree and hence has three roots. The equation can have positive/negative real roots as well as complex roots also.

**Step 3**:To determine the number of positive roots of a polynomial equation we have a rule known as Descartes’ Rule of Signs, which states that “The number of positive roots is at most equal to the number of sign changes. It can also be less than that by 2, 4, … i.e. if there are k sign changes in f(x), the number of positive roots could be k, k-2, and k-4,.....”

The coefficient of x^{3} is 1 which has a positive sign. The next term is -12x^{2 }and its coefficient is -12 and that has a negative sign so the first sign change takes place from first to the second term.

The third term is -25x and the coefficient is -25 which is negative. So when we move from the second term to the third term there is no sign change but when we move from the third term to the fourth term i.e. 5 there is a sign change.

The total number of sign changes in the given polynomial is 2 hence the number of positive roots is either 2 or zero. If two roots are positive then the third root has to be a conjugate complex root, but conjugate complex roots have to necessarily occur in pairs.

Hence two out of the three roots of the equation are complex conjugates and the third root is real but not positive.

## How many roots do the following equations have? -12x^{2 }- 25x + 5 + x^{3 }= 0

**Summary:**

The total number of roots of the equation f(x) = -12x^{2 }- 25x + 5 + x^{3 }= 0 is 3, of which two roots are complex and conjugate while the third root is non positive. Hence the equation has no positive real roots.

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