What are the Zeros of the Quadratic Function f(x) = 2x^2 – 10x – 3?

In the previous article, we have learned the ‘4 Best methods to find the Zeros of a Quadratic Function‘.

Now we understand this concept with the following example.

What are the zeros of the quadratic function f(x) = 2x^2 – 10x – 3?

Answer: x=\frac{5+\sqrt[]{31}}{2},\frac{5-\sqrt[]{31}}{2}

Detailed Solution:

The given quadratic function is f(x) = 2x^2 – 10x – 3 .

The zeros of the quadratic function f(x) = 2x^2 – 10x – 3 are the values of x for which the value of the quadratic function f(x) = 2x^2 – 10x – 3 is equal to zero.

Equating f(x) = 2x^2 – 10x – 3 with 0 , we get the following equation,

2x^2 – 10x – 3 = 0 …..\left ( 1 \right ) .

By the Formula of the Quadratic Equation, we know if ax^2 + bx + c = 0 …..\left ( 2 \right ), then the zeros are

x= \frac{-b\pm \sqrt{b^{2}-4ac} }{2a}…..\left ( 3 \right )

Comparing equations (1) and (2) we get,

a = 2, b = - 10, c = - 3 .

Putting these values in equation (3) we get,

x = \frac{ - \left ( - 10 \right ) \pm \sqrt{\left ( - 10 \right )^{2} - 4 \times 2 \times \left ( - 3 \right )}}{2 \times 2}

or, x = \frac{ 10 \pm \sqrt{ 100 + 24 }}{ 4 }

or, x = \frac{ 10 \pm \sqrt{ 124 }}{ 4 }

or, x = \frac{ 10 \pm 2 \sqrt{ 31 }}{ 4 }

or, x = \frac{ 5 \pm \sqrt{ 31 }}{ 2 }

or, x = \frac{ 5 + \sqrt{ 31 }}{ 2 }, \frac{ 5 - \sqrt{ 31 }}{ 2 }

Therefore the zeros of the quadratic function f(x) = 2x^2 – 10x – 3 are x = \frac{ 5 + \sqrt{ 31 }}{ 2 } and x = \frac{ 5 - \sqrt{ 31 }}{ 2 } .


Further Reading:

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