Question

x2-4x-3=0 solve this quadratic equation

Answer 1

x²-4x-3=0,

x=[4+-√{(-4)²-4×1×(-3)}]/2,

x=[4+-√{16+12}]/2,

x=[4+-√(28)]/2,

x=[4+-4√7]/2,

x=[2+-2√7],

then
x=(2+2√7),
and
x=(2-2√7)

Answer 2

Answer:

The roots of the equation are 4+√7  and 4-√7.

Step-by-step explanation:

Given: Quadratic equation x²-4x-3=0

To find: The values of x.

Solution: A  quadratic equation is a polynomial equation in a single variable where the highest exponent of the variable is 2.

A quadratic equation is a second order equation written in the form ax²+bx+c=0 where a,b, and c are coefficients of real numbers and a≠0.

The values of x that satisfy the equation are called solutions of the equation, and roots or zeros of the expression.

The quadratic equation can be solved by  three methods: factorising, using the quadratic formula, and completing the square.

A quadratic equation generally has two roots, which can be equal to each other. There are three cases

1. if the discriminant b²-4ac is positive, the roots are real and unequal.

2. if the discriminant is equal to zero, the two roots will be real and equal to each other.

3. if the discriminant is negative, the roots will be complex and each root is the complex conjugate of the other root.

Here the given quadratic expression cannot be easily factored , so we apply the formula.

x = -b ± $\frac{1}{2a} \sqrt{(b^{2}-4ac) }$

x²- 4x - 3= 0, here a= 1, b= -4, c= -3.

x  = -(-4)± $\frac{1}{2*1}$$\sqrt{((-4)^{2}-4*1*(-3) )}$

  = 4±$\frac{1}{2}$$\sqrt{(16+12)}$

  = 4±$\frac{1}{2} \sqrt{28}$

  = 4±$\frac{1}{2}$$\sqrt{4*7}$

  = 4±$\frac{1}{2}$ 2$\sqrt{7}$

x = 4+$\frac{2\sqrt{7} }{2}$ or x= 4 - $\frac{2\sqrt{7} }{2}$

x= 4+$\sqrt{7}$ or x= 4-$\sqrt{7}$ are the two values of x.

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