What values of b satisfy 3(26 + 3)² = 36?

Mathematics

1 answer:

MakcuM [25]1 year ago

7 0

The numeric value of the b is 0.23 which satisfy the given equation.

According to the statement

We have to find that the value of the b.

So, For this purpose, we know that the

The numeric value refers to the worth of each digit depending on where it lies in the number.

From the given information:

The equation is a

3(2b + 3)² = 36

then to find the value of b rearrange the terms then

(2b + 3)² = 12

(2b + 3) = \sqrt{12}

(2b) = \sqrt{12} -3

then

b = \sqrt{12} -3/ 2

b = 3.46 - 3/2

b = 0.46/2

b = 0.23.

So, The numeric value of the b is 0.23 which satisfy the given equation.

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The given question is a quadratic equation and we can use several methods to get the solutions to this question. The solution to the equation are 3/4 and -5/6 and the greater of the two solutions is 3/4

<h3>Quadratic Equation</h3>

Quadratic equation are polynomials with a second degree as it's highest power.

An example of a quadratic equation is

$y = ax^2 + bx + c$

The given quadratic equation is $24x^2 + 2x = 15$

Let's rearrange the equation

$24x^2 + 2x = 15\\24x^2 + 2x - 15 = 0$

This implies that

a = 24
b = 2
c = -15

The equation or formula of quadratic formula is given as

$y = \frac{-b +- \sqrt{b^2 - 4ac} }{2a}$

We can substitute the values into the equation and solve

$y = \frac{-b +- \sqrt{b^2 - 4ac} }{2a}\\y = \frac{-2 +- \sqrt{2^2 -4 * 24 * (-15)} }{2*24} \\y = \frac{-2+-\sqrt{4+1440} }{48} \\y = \frac{-2+-\sqrt{1444} }{48} \\y = \frac{-2+- 38}{48} \\y = \frac{-2+38}{48} \\y = \frac{3}{4}\\ \\or\\y = \frac{-2-38}{48} \\y = \frac{-40}{48} \\y = -\frac{5}{6}$