Respuesta :
Answer:
[tex]x=\frac{50+\sqrt{31}}{2},\frac{50-\sqrt{31}}{2}[/tex] are zeroes of given quadratic equation.
Step-by-step explanation:
We have been a quadratic equation:
[tex]2x^2-10x-3[/tex]
We need to find the zeroes of quadratic equation
We have a formula to find zeroes of a quadratic equation:
[tex]x=\frac{b^2\pm\sqrt{D}}{2a}\text{where}D=\sqrt{b^2-4ac}[/tex]
General form of quadratic equation is [tex]ax^2+bx+c[/tex]
On comparing general equation with b given equation we get
a=2,b=-10,c=-3
On substituting the values in formula we get
[tex]D=\sqrt{(-10)^2-4(2)(-3)}[/tex]
[tex]\Rightarrow D=\sqrt{100+24}=\sqrt{124}[/tex]
Now substituting D in [tex]x=\frac{b^2\pm\sqrt{D}}{2a}[/tex] we get
[tex]x=\frac{(-10)^2\pm\sqrt{124}}{2\cdot 2}[/tex]
[tex]x=\frac{100\pm\sqrt{124}}{4}[/tex]
[tex]x=\frac{100\pm2\sqrt{31}}{4}[/tex]
[tex]x=\frac{50\pm\sqrt{31}}{2}[/tex]
Therefore, [tex]x=\frac{50+\sqrt{31}}{2},\frac{50-\sqrt{31}}{2}[/tex]
Answer:
A quadratic equation is in the form of [tex]ax^2+bx+c = 0[/tex] .......[1]
then the solution for this equation is given by:
[tex]x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}[/tex]
Given the quadratic equation:
[tex]f(x) = 2x^2-10x-3[/tex]
To find the zero of the given equation.
Set f(x) = 0
then;
[tex]2x^2-10x-3=0[/tex]
On comparing with equation [1] we have;
a = 2, b = -10 and c = -3
then;
[tex]x = \frac{-(-10) \pm \sqrt{(-10)^2-4(2)(-3)}}{2(2)}[/tex]
⇒[tex]x = \frac{10 \pm \sqrt{100+24}}{4}[/tex]
⇒[tex]x = \frac{10 \pm \sqrt{124}}{4}[/tex]
⇒[tex]x = \frac{10 \pm 2\sqrt{31}}{4}[/tex]
Simplify:
[tex]x = \frac{5 \pm \sqrt{31}}{2}[/tex]
Therefore, the zeros of the given quadratic equation are;
[tex]x = \frac{5+\sqrt{31}}{2}[/tex] and [tex]x=\frac{5-\sqrt{31}}{2}[/tex]
