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Dima020 [189]
2 years ago
6

Harry is trying to solve the equation 0 = 2x2 − x − 6 using the quadratic formula. He has made an error in one of the steps belo

w. Find the step where Harry went wrong. (1 point)
Mathematics
1 answer:
bija089 [108]2 years ago
4 0
<h3><u>Given</u> - </h3>

➙ a quadratic equation in which Harry lagged due to an error made by him, 2x² - x - 6= 0

<h3><u>To solve</u> - </h3>

➙ the given quadratic equation.

<h3><u>Concept applied</u> - </h3>

➙ We will apply the quadratic formula as given in the question. So, let's study about quadratic equation first because we are supposed to apply the formula in equation.

What is quadratic equation?

➙ A quadratic equation in the variable x is an equation of the form ax² + bx + c = 0, where a, b, c are real numbers, a ≠ 0.

Now, what is quadratic formula?

➙The roots of a quadratic equation ax + bx + c = 0 are given by \sf{\:\frac{-b \pm\: \sqrt {b ^ 2 - 4ac}}{2a}} provided b - 4ac ≥ 0.

<h3><u>Solution</u> - </h3>

here as per the given quadratic equation,

a = 2, b = -1 and c = -6

putting in the formula,

\implies\sf{x=\frac{-(-1) \pm\: \sqrt {(-1)^2 - 4(2)(-6)}}{2(2)}}

\implies\sf{x=\frac{1 \pm\: \sqrt {1+48}}{4}}

\implies\sf{x=\frac{1 \pm\: \sqrt {49}}{4}}

\implies\sf{x=\frac{1 \pm\: 7}{4}}

Solving one by one,

\implies\sf{x=\frac{1 + \: 7}{4}}

\implies\sf{x=\frac{8}{4}}

\implies{\boxed{\bf{x=2}}}

________________

\implies\sf{x=\frac{1 - \: 7}{4}}

\implies\sf{x=\frac{-6}{4}}

\implies{\boxed{\bf{x=\frac{-3}{2}}}}

________________________________

<em><u>Note</u> - Hey dear user!! You haven't provided the solution which was solved by Harry (A.T.Q). Please go through the solution as it will help you to find the error done by Harry.</em>

<em>________________________________</em>

Hope it helps!! (:

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The area of the region which is inside the polar curve r = 5 sinθ but outside r = 4 will be 3.75 square units.

<h3>What is an area bounded by the curve?</h3>

When the two curves intersect then they bound the region is known as the area bounded by the curve.

The area of the region which is inside the polar curve r = 5 sinθ but outside r = 4 will be

Then the intersection point will be given as

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Then by the integration, we have

\rightarrow \dfrac{1}{2} \times \int _{0.927}^{2.214}[ (5 \sin \theta)^2 - 4^2] d\theta \\\\\\\rightarrow \dfrac{1}{2} \times \int _{0.927}^{2.214} [25\sin ^2 \theta - 16] d\theta \\\\\\\rightarrow \dfrac{1}{2} \times \int _{0.927}^{2.214} [ \dfrac{25}{2}(1 - \cos 2\theta ) - 16] d\theta \\

\rightarrow \dfrac{1}{2} [\dfrac{25 \theta }{5} - \dfrac{25 \cos 2\theta }{2} - 16\theta]_{0.927}^{2.214} \\\\\\\rightarrow \dfrac{1}{2} [\dfrac{25(2.214 - 0.927) }{5} - \dfrac{25 (\cos 2\times 2.214 - \cos 2\times 0.927) }{2} - 16(2.214 - 0.927]\\

On solving, we have

\rightarrow \dfrac{1}{2} \times 7.499\\\\\rightarrow 3.75

Thus, the area of the region is 3.75 square units.

More about the area bounded by the curve link is given below.

brainly.com/question/24563834

#SPJ4

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