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2 years ago

Use the number line below, where RS = 5y + 2, ST = 3y + 6, and RT = 48

Mathematics

1 answer:

tekilochka [14]2 years ago

3 0

RS+ST= RT

(7Y+3)+(2Y+9)= 14Y-8

9Y+12=14Y-8

5Y=20

Y=4

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Find the number of real number solutions for the equation. x^2 + 5x + 7 = 0

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The number of real solutions for the equation $x^{2}+5 x+7=0$ is zero

Solution:

For a Quadratic Equation of form : $a x^{2}+b x+c=0$ ---- eqn 1

The solution is $x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

Now , the given Quadratic Equation is $x^{2}+5 x+7=0$ ---- eqn 2

On comparing Equation (1) and Equation(2), we get

a = 1 , b = 5 and c = 7

In $x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$ , $b^2 - 4ac$ is called the discriminant of the quadratic equation

Its value determines the nature of roots

Now, here are the rules with discriminants:

1) D > 0; there are 2 real solutions in the equation

2) D = 0; there is 1 real solution in the equation

3) D < 0; there are no real solutions in the equation

Now let solve for given equation

$D= b^2 - 4ac\\\\D = 5^2 - 4(1)(7)\\\\D = 25 - 28 \\\\D = -3$

Since -3 is less than 0, this means that there are 0 real solutions in this equation.

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Use the number line​ below, where RS = 5y + 2, ST = 3y + 6, and RT = 48

Use the number line below, where RS = 5y + 2, ST = 3y + 6, and RT = 48