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Chapter 7: Problem 19

Solve the following quadratic equations by an appropriate method. (a) \(x^{2}+16 x+64=0\) (b) \(x^{2}-6 x+3=0\) (c) \(2 x^{2}-6 x-3=0\) (d) \(x^{2}-4 x+1=0\) (e) \(x^{2}-22 x+121=0\) (f) \(x^{2}-8=0\)

Short Answer

Expert verified
Question: Solve the given quadratic equations and find the solutions for x. (a) \(x^2 + 16x +64 = 0\) (b) \(x^2 - 6x + 3 = 0\) (c) \(2x^2 - 6x - 3 = 0\) (d) \(x^2 - 4x + 1 = 0\) (e) \(x^2 - 22x + 121 = 0\) (f) \(x^2 - 8 = 0\) Answer: (a) x = -8 (b) \(x = 3 +\frac{\sqrt{24}}{2}\), \(x = 3 - \frac{\sqrt{24}}{2}\) (c) \(x = \frac{3 + \sqrt{15}}{2}\), \(x = \frac{3 - \sqrt{15}}{2}\) (d) \(x = 2 + \frac{\sqrt{12}}{2}\), \(x = 2 - \frac{\sqrt{12}}{2}\) (e) x = 11 (f) x = \(\sqrt{8}\), x = \(-\sqrt{8}\)

Step by step solution

01

(a) Factoring the quadratic equation

The equation is given as: \(x^{2}+16 x+64=0\). We can factor this equation as \((x+8)(x+8)=0\), and therefore, find the solution for x.
02

(a) Finding the solution for x

We have factored the equation as \((x+8)(x+8)=0\). If any of the factors are equal to 0, the whole equation will be equal to 0. Therefore, we can say: \(x+8=0\) or \(x+8=0\) So, x = -8 is the solution.
03

(b) Factoring the quadratic equation

The equation is given as: \(x^{2}-6 x+3=0\). This equation is not easy to factor. So, we will solve it using the quadratic formula.
04

(b) Applying quadratic formula

The equation is \(x^{2}-6x+3=0\). Comparing it with the general form \(ax^2+bx+c=0\), we get \(a=1\), \(b=-6\), and \(c=3\). Using the quadratic formula, we find the solutions for x: \[x=\frac{-(-6)\pm\sqrt{(-6)^2-4(1)(3)}}{2(1)}\] \[x=\frac{6\pm\sqrt{36-12}}{2}\] \[x=\frac{6\pm\sqrt{24}}{2}\] \[x=3\pm\frac{\sqrt{24}}{2}\] So, the solutions are \(x=3+\frac{\sqrt{24}}{2}\) and \(x=3-\frac{\sqrt{24}}{2}\).
05

(c) Applying quadratic formula

The equation is \(2x^{2}-6x-3=0\). Comparing it with the general form \(ax^2+bx+c=0\), we get \(a=2\), \(b=-6\), and \(c=-3\). Now, apply the quadratic formula to find the solutions for x: \[x=\frac{-(-6)\pm\sqrt{(-6)^2-4(2)(-3)}}{2(2)}\] \[x=\frac{6\pm\sqrt{36+24}}{4}\] \[x=\frac{6\pm\sqrt{60}}{4}\] So, the solutions are \(x=\frac{3+\sqrt{15}}{2}\) and \(x=\frac{3-\sqrt{15}}{2}\).
06

(d) Applying quadratic formula

The equation is \(x^{2}-4x+1=0\). Comparing it with the general form \(ax^2+bx+c=0\), we get \(a=1\), \(b=-4\), and \(c=1\). Now, apply the quadratic formula to find the solutions for x: \[x=\frac{-(-4)\pm\sqrt{(-4)^2-4(1)(1)}}{2(1)}\] \[x=\frac{4\pm\sqrt{16-4}}{2}\] \[x=\frac{4\pm\sqrt{12}}{2}\] So, the solutions are \(x=2+\frac{\sqrt{12}}{2}\) and \(x=2-\frac{\sqrt{12}}{2}\).
07

(e) Factoring the quadratic equation

The equation is given as: \(x^{2}-22x+121=0\). We can factor this equation as \((x-11)(x-11)=0\), and therefore, find the solution for x.
08

(e) Finding the solution for x

We have factored the equation as \((x-11)(x-11)=0\). If any of the factors are equal to 0, the whole equation will be equal to 0. Therefore, we can say: \(x-11=0\) or \(x-11=0\) So, x = 11 is the solution.
09

(f) Factoring the quadratic equation

The equation is given as: \(x^{2}-8=0\). We can factor this equation as \((x-\sqrt{8})(x+\sqrt{8})=0\), and therefore, find the solution for x.
10

(f) Finding the solution for x

We have factored the equation as \((x-\sqrt{8})(x+\sqrt{8})=0\). If any of the factors are equal to 0, the whole equation will be equal to 0. Therefore, we can say: \(x-\sqrt{8}=0\) or \(x+\sqrt{8}=0\) So, the solutions are x = \(\sqrt{8}\) and x = \(-\sqrt{8}\).

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Most popular questions from this chapter

On a number line show the numbers \(-\pi, 0\), \(\sqrt{2},-\sqrt{3},|-0.5|,-(3 !)\) and \(\frac{11}{19}\)

If \(x^{3}-17 x^{2}+54 x-8=(x-4) \times\) a polynomial, state the degree of the polynomial.

Use the method of completing the square to derive the formula for solving a quadratic equation.

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