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1 year ago

Roots of Quadratics 50 PTS!!!

Mathematics

2 answers:

shusha [124]1 year ago

8 0

Answer:

ax² + bx + c = 0 is solved using the formula

Step-by-step explanation:

noname [10]1 year ago

4 0

The values of k for the different quadratic equation solutions are as follows

a the equation 2x² - x + 3k = 0 has two distinct real roots

k < 1/24

b. the equation 5x² - 2x + (2k − 1) = 0 has equal roots

k = 3/5

ci the equation -x² + 3x + (k + 1) = 0 has real roots

k > -3.25

d the equation 3kx² - 3x + 2 = 0 has no real solutions

k < ± 1.633

<h3>How to solve quadratic equations to get different answers</h3>

Quadratic equations of the form ax² + bx + c = 0 is solved using the formula

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

The equation b² - 4ac is called the discriminant and it is used as follows

To solve the equation and get two real roots: 2x² - x + 3k = 0

b² - 4ac > 0

substituting the values gives

(-1)² - 4 * 2 * 3k > 0

1 - 24k > 0

1 > 24k

divide through by coefficient of k

k < 1/24

To solve the equation and get equal roots: 5x² - 2x + (2k − 1) = 0

b² - 4ac = 0

substituting the values gives

(-2)² - 4 * 5 * (2k - 1) = 0

4 - 40k + 20 = 0

-40k = -24

divide through by coefficient of k

k = 3/5

To solve the equation and get real roots -x² + 3x + (k + 1) = 0

b² - 4ac > 0

substituting the values gives

(3)² - 4 * -1 * (k+1) > 0

9 + 4k + 4> 0

4k > -13

divide through by coefficient of k

k > -3.25

To solve the equation and get no real solutions 3kx² - 3x + 2 = 0

b² - 4ac < 0

substituting the values gives

(-3)² - 4 * 3k * 2 < 0

9 - 24k² > 0

9 > 24k²

divide through by coefficient of k²

k² < 24/9

k < ± 1.633

Learn more about roots of quadratic equations: brainly.com/question/26926523

#SPJ1

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These are 8 questions and 8 answers:

1) Quesion 1:

9+√2
---------
4 - √7

Answer: the third option:

36 + 9√7 + 4√2 + √14
-----------------------------
9

Explanation:

Multiply both numerator and denominator by the conjugate of the denominator.

The conjugate of 4 - √7 = 4 + √7

=>

$\frac{9+ \sqrt{2} }{4- \sqrt{7} } . \frac{4+ \sqrt{7} }{4+ \sqrt{7} } = \frac{(9)(4)+9 \sqrt{7}+4 \sqrt{2} + \sqrt{2} . \sqrt{7} }{(4)^2-( \sqrt{7})^2 } =$

$= \frac{36+9 \sqrt{7} +4 \sqrt{2} + \sqrt{14} }{16-7}$

2) Question 2: sum

$5x (\sqrt[3]{x^2y})+2( \sqrt[3]{x^5y})$

Answer: fourth option

$7x( \sqrt[3]{x^2y} )$

Explanation:

Take x^5 out of the second radical which will result in a like term of the first radical:

$5x( \sqrt[3]{x^2y} )+2( \sqrt[3]{x^5y}) =5x( \sqrt[3]{x^2y} )+2x( \sqrt[3]{x^2y})=7x( \sqrt[3]{x^2y})$

which is the fourth option

3) Question 3. Which expression is equivalent to:

$\frac{ \sqrt{10} }{ \sqrt[4]{8} }$

Answer: the first option

Explanation

$\frac{ \sqrt{10} }{ \sqrt[4]{8} } = \frac{ \sqrt[4]{10^2} }{ \sqrt[4]{8} } = \frac{ \sqrt[4]{100} }{ \sqrt[4]{8} } . \frac{ \sqrt[4]{8^3} }{ \sqrt[4]{8^3} } = \frac{ \sqrt[4]{(100)(512)} }{8} = \frac{ \sqrt[4]{51200} }{8} = \frac{4 \sqrt[4]{200} }{8} = \frac{ \sqrt[4]{200} }{2}$

4) Question 4 What is the simplest form?

Answer: the second option

Explanation:

$\sqrt[4]{81x^8y^5}=x^2 y\sqrt[4]{3^4y} =3x^2y \sqrt[4]{y}$

5) Question 5 Product

Answer: the fourth option:

$104x^4+16x^4 \sqrt{30} [/tex]\\Explanation:\\Use the square of a binomial product: (a + b)^2 = a^2 + 2ab + b^2\\[tex](4x \sqrt{5x^2} )^2+2(4x \sqrt{5x^2})(2x^2 \sqrt{6}) +(2x^2 \sqrt{6} )^2=$

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

$=104x^4+16x^4 \sqrt{30}$

which is the fourth option.

6) Question 6 Product

Answer: fourth option

Explanation:

$\sqrt[3]{16x^7} . \sqrt[3]{12x^9} = \sqrt[3]{2^4.2^2.3x^7x^9} = \sqrt[3]{2^6.3.x^{16}}=2^2 x^5 \sqrt[3]{3x} =4 x^5\sqrt[3]{3x}$

which is the fourth option.

7) Question 7. Simplified form of 2√18 + 3√2 + √162

Answer: 18√2

Explanation:

$2 \sqrt{18}+3 \sqrt{2} + \sqrt{162}=2(3) \sqrt{2} + 3 \sqrt{2} +9 \sqrt{2} =18 \sqrt{2}$

which is the second option.

8) Question 8 which function is undefined for x = 0.

Answer: second option y = √ (x - 2)

Explanation.

The square root function is not defined for negative values.

When x = 0, x - 2 = -2, whose square root is not defined.

Therefore, the square root of x - 2 is not defined for x = 0.