Question

How many real-number solutions does the equation have?

Answer 1

Answer:

Option b. Two solutions

Step-by-step explanation:

In order to find how many real number solutions the equation has we have to solve it

Given equation: -4x² + 10x + 6 = 0

taking 2 common from the equation

2(-2x² + 5x + 3) = 0

-2x² + 5x + 3 = 0

taking minus sign common from the above equation

2x² - 5x - 3 = 0

We will solve this equation by factorization in such a way that the sum of two factors is equal to -5x and the product is -6x²

2x² - 6x + x - 3 = 0

taking common above

2x(x-3) + 1(x-3) = 0

taking (x-3) common

(2x+1)(x-3) = 0

2x + 1 = 0

2x = -1

x = $\frac{-1}{2}$

x - 3 = 0; x = 3

the solutions are

$x=\frac{-1}{2},3$

Both values are real numbers, therefore correct option is b

Answer 2

Answer:

Choice b is correct answer.

Step-by-step explanation:

Given equation is :

-4x²+10x+6 = 0

ax²+bx+c = 0 is general quadratic equation.

comparing general equation with given quadratic equation,we get

a = -4 ,b = 10 and c = 6

we use the discriminant formula to find number of solution.

If D > 0 then there is two real solutions.

The formula to find discriminant is :

D = b²-4ac

putting above values in discriminant formula, we get

D = (10)²-4(-4)(6)

D = 100+96

D = 196 > 0

hence , equation have two real equation.