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lisov135 [29]
3 years ago
7

What are the zeros of the quadratic function f(x) = 2x2 – 10x – 3?

Mathematics
2 answers:
Maslowich3 years ago
7 0

Answer:

A quadratic equation is in the form of ax^2+bx+c = 0 .......[1]

then the solution for this equation is given by:

x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}

Given the quadratic equation:

f(x) = 2x^2-10x-3

To find the zero of the given equation.

Set f(x) = 0

then;

2x^2-10x-3=0

On comparing with equation [1] we have;

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

then;

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

⇒x = \frac{10 \pm \sqrt{100+24}}{4}

⇒x = \frac{10 \pm \sqrt{124}}{4}

⇒x = \frac{10 \pm 2\sqrt{31}}{4}

Simplify:

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

Therefore, the zeros of the given quadratic equation are;

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

Natasha2012 [34]3 years ago
5 0

Answer:

x=\frac{50+\sqrt{31}}{2},\frac{50-\sqrt{31}}{2}  are zeroes of given quadratic equation.

Step-by-step explanation:

We have been a quadratic equation:

2x^2-10x-3

We need to find the zeroes of quadratic equation

We have a formula to find zeroes of a quadratic equation:

x=\frac{b^2\pm\sqrt{D}}{2a}\text{where}D=\sqrt{b^2-4ac}

General form of quadratic equation is ax^2+bx+c

On comparing general equation with b given equation we get

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

On substituting the values in formula we get

D=\sqrt{(-10)^2-4(2)(-3)}

\Rightarrow D=\sqrt{100+24}=\sqrt{124}

Now substituting D in  x=\frac{b^2\pm\sqrt{D}}{2a} we get

x=\frac{(-10)^2\pm\sqrt{124}}{2\cdot 2}

x=\frac{100\pm\sqrt{124}}{4}

x=\frac{100\pm2\sqrt{31}}{4}

x=\frac{50\pm\sqrt{31}}{2}

Therefore, x=\frac{50+\sqrt{31}}{2},\frac{50-\sqrt{31}}{2}



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