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3 years ago

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Mathematics

1 answer:

Bond [772]3 years ago

5 0

Answer:

Below

Step-by-step explanation:

The equation is x2 + bx + c

The coefficients are

● a = 1

● b

● c is the constant

The difference of the roots is | b - 2c |

Let s and p be the roots of the polynomial

● s - p = | b - 2c | (1)

● s + p = -b/a => s + p = -b (2)

(1) + (2)

● s - p + s + p = |b-2c| - b

● 2s = | b - 2c | - b

● 2s = b - 2c - b or 2s = 2c - b - b

● 2s = - 2c or 2s = 2c - 2b

● s = - c or s = c - b

We want to express b in term of c. So we will be intetsted in the 2nd equation

● s = c - b

● s - c = - b

● b = c - s

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Read 2 more answers

What is the following product? <br>(4x square root 5x^2 + 2^2 square root 6)^2

tangare [24]

The product is $104 x^{4}+16 \sqrt{30} x^{4}$

Explanation:

The given expression is $\left(4 x \sqrt{5 x^{2}}+2 x^{2} \sqrt{6}\right)^{2}$

We need to determine the product of the given expression.

First, we shall simplify the given expression.

Thus, we have,

$\left(4 x \sqrt{5 x^{2}}+2 x^{2} \sqrt{6}\right)^{2}=\left(4 x \sqrt{5} x+2 x^{2} \sqrt{6}\right)^2$

$\left(4 x \sqrt{5 x^{2}}+2 x^{2} \sqrt{6}\right)^{2}=\left(4 x^{2} \sqrt{5}+2 x^{2} \sqrt{6}\right)^2$

Expanding the expression, we have,

$\left(4 x \sqrt{5 x^{2}}+2 x^{2} \sqrt{6}\right)^{2}=\left(4 x^{2} \sqrt{5}+2 x^{2} \sqrt{6}\right)\left(4 x^{2} \sqrt{5}+2 x^{2} \sqrt{6}\right)$

Now, we shall apply FOIL, we get,

$\left(4 x \sqrt{5 x^{2}}+2 x^{2} \sqrt{6}\right)^{2}=\left(4 x^{2} \sqrt{5}\right)^{2}+2 ( 2 x^{2} \sqrt{6})(4 x^{2} \sqrt{5})+\left(2 x^{2} \sqrt{6}\right)^{2}$

Simplifying the terms, we have,

$\left(4 x \sqrt{5 x^{2}}+2 x^{2} \sqrt{6}\right)^{2}=16 \cdot 5 x^{4}+16 \sqrt{30} x^{4}+4 \cdot 6 x^{4}$

Multiplying, we get,

$\left(4 x \sqrt{5 x^{2}}+2 x^{2} \sqrt{6}\right)^{2}=80 x^{4}+16 \sqrt{30} x^{4}+24 x^{4}$

Adding the like terms, we get,

$\left(4 x \sqrt{5 x^{2}}+2 x^{2} \sqrt{6}\right)^{2}=104 x^{4}+16 \sqrt{30} x^{4}$

Thus, the product of the given expression is $104 x^{4}+16 \sqrt{30} x^{4}$

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3 years ago