Question

show that the equation 3x² + 7x + 8 = 0is not true for any real value of'x'​

Answer 1

$\huge\blue{\boxed{\underline{ANSWER}}}$

GIVEN THAT;

$&#10156$ Quadratic equation 3x² + 7x + 8 = 0

SOLUTIONS

$&#10156$ The standard form of quadratic equation is ax² + bx + c = 0

Now comparing the given equation with standard equaton

a = 3

b = 7

c = 8

If the value of D is less than zero then the equation is not true for any real value of x

$&#10156$ D = b² - 4ac

$&#10230 \: \: D \: = {b}^{2} - 4ac \: \: \: \: \: \: \: \: \: \: \\ \\ &#10230 \: \: D = {7}^{2} - 4 \times 3 \times 8 \\ \\ &#10230 \: \: D = 49 - 96 \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ &#10230 \: \: D = - 47 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

• The value of D is a less than 0 so this equation is not true for any real numbers of x .

Answer 2

Answer:

Step-by-step explanation:

To Show:-

3x^2+7x+8 = 0 is not true for any real value of 'x'.

Solution :-

To show that , first we need to know about the conditions of Discriminant(Delta) :-

Consider a General Quadratic equation :-

ax^2 + bx + c = 0.

Discriminant Δ for the above equation is  = b^2 - 4ac

1) If Δ > 0 Then the roots of the equation  is real and distinct.

2) If Δ = 0 Then the roots of the equation  is real and equal.

3) If Δ < 0 Then the roots of the equation is complex and conjugate to each other.[No real roots]

Let's do :-

Comparing "3x^2+7x+8 = 0" with general form of quadratic equation "ax^2+bx+c = 0" :-

We can say that :-

a = 3 , b = 7 , c = 8

Δ = b^2 - 4ac

= (7)^2 - 4(3)(8)

= 49 - 96

= -47

Δ= -47 < 0

Since according to 3rd condition of discriminant :- the equation 3x² + 7x + 8 = 0is not true for any real value of  'x'.