Class 8 question
Find the value of x using quadratic formula
16x^4- 28x² - 8=0
Substitute x² with t
16 x⁴ - 28 x² - 8 = 0
become
16 t² - 28 t - 8 = 0
Divide both sides by 4
4 t² - 7 t - 2 = 0
The coefficients are:
a = 4 , b = - 7 , c = - 2
t1/2 = [ - b ± √ ( b² - 4 a c ) ] / 2 a =
[ - ( - 7 ) ± √ ( ( - 7 )² - 4 ∙ 4 ∙ ( - 2 ) ) ] / 2 ∙ 4 =
[ 7 ± √ ( 49 + 32 ) ] / 8 = ( 7 ± √81 ) / 8 = ( 7 ± 9 ) / 8
t1 = ( 7 + 9 ) / 8 = 16 / 8 = 2
t2 = ( 7 - 9 ) / 8 = - 2 / 8 = - 1 / 4
If x² = t
then
x = ± √ t
t1 = 2
x = ± √2
t2 = - 1 / 4
x = ± √ -1 / √4 = ± i / 2
The solutions are:
x = - i / 2 , x = i / 2 , x = -√2 , x = -√2
The solutions are:
x = - i / 2 , x = i / 2 , x = -√2 , x = √2
To find the value of x using the quadratic formula, you need to have a quadratic equation in the form of ax^2 + bx + c = 0. In this case, the given equation is 16x^4 - 28x^2 - 8 = 0.
Step 1: Identify the coefficients of the quadratic equation.
In our equation, a = 16, b = -28, and c = -8.
Step 2: Substitute the values of a, b, and c into the quadratic formula:
The quadratic formula is given by: x = (-b ± √(b^2 - 4ac)) / 2a
Substituting the values, we have:
x = (-(-28) ± √((-28)^2 - 4 * 16 * (-8))) / (2 * 16)
x = (28 ± √(784 + 512)) / 32
x = (28 ± √1296) / 32
Step 3: Simplify the expression:
x = (28 ± 36) / 32
This gives two possible values for x:
x1 = (28 + 36) / 32 = 64 / 32 = 2
x2 = (28 - 36) / 32 = -8 / 32 = -1/4
Therefore, the values of x that satisfy the quadratic equation 16x^4 - 28x^2 - 8 = 0 are x = 2 and x = -1/4.