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Alisiya [41]
4 years ago
11

What method would you choose to solve the equation 2x2 – 7 = 9? Explain why you chose this method. 

Mathematics
2 answers:
horsena [70]4 years ago
7 0

Answer:

I would use the square root property of equality. Because there is no x term, I would just add and divide to get x2 by itself. Then I would take the square root.

lisabon 2012 [21]4 years ago
4 0

is that a variable of x or a multiplication sign?

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Answer:

\displaystyle \left(\alpha+1\right)\left(\beta + 1\right)  = \frac{a+c-b}{a}\:\: \left(\text{ or } 1+\frac{c-b}{a}\right)

Step-by-step explanation:

We are given the equation:

ax^2+bx+c=0

Which has roots α and β.

And we want to express (α + 1)(β + 1) in terms of <em>a</em>, <em>b</em>, and <em>c</em>.

From the quadratic formula, we know that the two solutions to our equation are:

\displaystyle x_1 = \frac{-b+\sqrt{b^2-4ac}}{2a}\text{ and } x_2=\frac{-b-\sqrt{b^2-4ac}}{2a}

Let <em>x</em>₁ = α and <em>x₂ </em>= β. Substitute:

\displaystyle \left(\frac{-b+\sqrt{b^2-4ac}}{2a} + 1\right) \left(\frac{-b-\sqrt{b^2-4ac}}{2a}+1\right)

Combine fractions:

\displaystyle =\left(\frac{-b+2a+\sqrt{b^2-4ac}}{2a} \right) \left(\frac{-b+2a-\sqrt{b^2-4ac}}{2a}\right)

Rewrite:

\displaystyle = \frac{\left(-b+2a+\sqrt{b^2-4ac}\right)\left(-b+2a-\sqrt{b^2-4ac}\right)}{(2a)(2a)}

Multiply and group:

\displaystyle = \frac{((-b+2a)+\sqrt{b^2-4ac})((-b+2a)-\sqrt{b^2-4ac})}{4a^2}

Difference of two squares:

\displaystyle = \frac{\overbrace{(-b+2a)^2 - (\sqrt{b^2-4ac})^2}^{(x+y)(x-y)=x^2-y^2}}{4a^2}

Expand and simplify:

\displaystyle = \frac{(b^2-4ab+4a^2)-(b^2-4ac)}{4a^2}

Distribute:

\displaystyle = \frac{(b^2-4ab+4a^2)+(-b^2+4ac)}{4a^2}

Cancel like terms:

\displaystyle = \frac{4a^2+4ac-4ab}{4a^2}

Factor:

\displaystyle =\frac{4a(a+c-b)}{4a(a)}

Cancel. Hence:

\displaystyle = \frac{a+c-b}{a}\:\: \left(\text{ or } 1+\frac{c-b}{a}\right)

Therefore:

\displaystyle \left(\alpha+1\right)\left(\beta + 1\right) = \frac{a+c-b}{a}

4 0
3 years ago
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