Question

Help with these two problems and show work please !!

Answer 1

Answer:

1) $x_1=-1-2\sqrt{2},\ x_2=-1+2\sqrt{2}$

2) $x_1=\dfrac{-5 - \sqrt{13}}{6},\ x_2=\dfrac{-5 + \sqrt{13}}{6}$

Step-by-step explanation:

${\large \textsf{ Quadratic Formula: }}x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}\ \Bigg{\|}\ \textsf{when }ax^2+bx+c=0$

Given quadratic equations:

1. $x^2+2x-7=0$

2. $4x^2-3=x^2-5x-4$

1. x² + 2x - 7 = 0

$\implies a=\textsf{1},b=\textsf{2},c=\textsf{-7}$

Step 1: Substitute the given values into the formula and simplify.

$\begin{aligned}\implies x&=\dfrac{-(\textsf{2})\pm \sqrt{(\textsf{2})^2-4(\textsf{1})(\textsf{-7})}}{2(\textsf{1})}\\\implies x&=\dfrac{-2\pm \sqrt{4-4(-7)}}{2}\\\implies x&=\dfrac{-2\pm \sqrt{4+28}}{2}\\\implies x&=\dfrac{-2\pm \sqrt{32}}{2}\end{aligned}$

Step 2: Simplify the radicand (under the square root).

$\begin{aligned}x&=\dfrac{-2\pm \sqrt{16\times2}}{2}\\x&=\dfrac{-2\pm 4\times\sqrt{2}}{2}\\x&=\dfrac{-2\pm 4\sqrt{2}}{2}\end{aligned}$

Step 3: Separate into two solutions and simplify them.

$\implies x_1&=\dfrac{-2 - 4\sqrt{2}}{2},\ x_2&=\dfrac{-2 + 4\sqrt{2}}{2}$

$\begin{aligned}\implies x_1&=\dfrac{-2}{2}+\dfrac{- 4\sqrt{2}}{2},\ x_2=\dfrac{-2 + 4\sqrt{2}}{2}\\\implies {x_1&=\boxed{-1-2\sqrt{2}},\ x_2=\boxed{-1+2\sqrt{2}} \end{aligned}$

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2. 4x² - 3 = x² - 5x - 4

Step 1: Set the equation to zero (by moving the "x² - 5x - 4" to the left).

4x² - 3 - x² + 5x + 4 = 0 [ Combine like terms. ]

3x² + 5x + 4 = 0

$\implies a=\textsf{3},b=\textsf{5},c=\textsf{1}$

Step 2: Substitute the given values into the formula and simplify.

$\begin{aligned}\implies x&=\dfrac{-(\textsf{5})\pm \sqrt{(\textsf{5})^2-4(\textsf{3})(\textsf{1})}}{2(\textsf{3})}\\\implies x&=\dfrac{-5\pm \sqrt{25-12}}{6}\\\implies x&=\dfrac{-5\pm \sqrt{13}}{6}\end{aligned}$

Step 3: Separate into two solutions.

$\implies x_1=\dfrac{-5 - \sqrt{13}}{6},\ x_2=\dfrac{-5 + \sqrt{13}}{6}$

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