﻿ Solving Equations in Quadratic Form Worksheet | Problems & Solutions # Solving Equations in Quadratic Form Worksheet

Solving Equations in Quadratic Form Worksheet
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1.
Find the solution of the equation $x$4 - 11$x$2 + 30 = 0. a. {- 6, 6, - 5, 5} b. {- 2$\sqrt{6}$, 2$\sqrt{6}$, - 2$\sqrt{5}$, 2$\sqrt{5}$} c. {- 3$\sqrt{6}$, 3$\sqrt{6}$, - 2$\sqrt{6}$, 2$\sqrt{6}$} d. {- $\sqrt{6}$, $\sqrt{6}$, - $\sqrt{5}$, $\sqrt{5}$}

#### Solution:

x4 - 11x2 + 30 = 0

Let x2 = t, then the equation reduces to t2 - 11t + 30 = 0

t2 - 11t + 30 = 0

t2 - 6t - 5t + 30 = 0 = t2 - 6t - 5(t - 6) = 0

(t - 6) (t - 5) = 0 t = 6, t = 5

Therefore, the roots of t2 - 11t + 30 = 0 are 6 and 5.
[Equate each term to zero.]

x4 - 11x2 + 30 = 0
[Substitute t = x2 = 6 and 5.]

x2 = 6 or x2 = 5
[6 and 5 are roots of the equation.]

x = ± 6 or x = ± 5 x = + 6, - 6 or +5, - 5
[Square on both the sides.]

Therefore, the solution of the equation is {- 6, 6, - 5, 5}.

2.
The number of real solutions of the equation |$x$|2 - 3|$x$| + 2 = 0 is _____ a. 4 b. 1 c. 2 d. 3

#### Solution:

Let x be the positive number.

Then x2 - 3x + 2 = 0
[Substitute x as positive in the given equation.]

(x - 2) (x - 1) = 0
[Factor.]

x = 2, 1

Let x be the negative number.

Then x2 + 3x + 2 = 0
[Substitute x as negative in the given equation.]

(x + 2) (x + 1) = 0
[Factor.]

x = -2, -1

Hence there are 4 real solutions for the given equation.

3.
Solve the equation 2 $x$ + = 0 a. 4 b. 1 and $\frac{1}{4}$ c. $\frac{1}{4}$ d. 1

#### Solution:

The given equation is 2x + x - 1= 0

x = 1 - 2x
[Bring all the terms to one side except x.]

x = 1 - 4x + 4x2
[Squaring the above equation on both sides.]

4 x2 - 5 x + 1 = 0
[Write in the form of an equation ax2 + bx + c = 0.]

(x - 1) (4x - 1) = 0
[Factors.]

x = 1, 1 / 4
[Solve for x.]

When x =1, 1 = 1 - 2(1)
1 ≠ - 1
[Substitute the values in x = 1 - 2x.]

x = 1 does not satisfy the equation.
[From step 7.]

When x = 1 / 4, 14 = 1 - 1 / 2
[Check by substituting in x = 1 - 2x.]

1 / 4= 1 / 4, x = 1 / 4satisfies the equation.
[Squaring on both sides.]

Hence 1 / 4 is the only solution.

4.
The equation $x$2 + 143 = 43 has a solution in which set of numbers? a. Integers b. Irrational numbers c. Complex numbers d. Natural numbers

#### Solution:

x2 + 143 = 43

x2 = 43 - 143
[Subtract 43 to each side of the equation.]

x2 = - 100
[Simplify.]

x = ±- 100

x = ±10i, which are complex numbers.

So, the equation x2 + 143 = 43 has a solution in complex numbers.

5.
Verify if the equation can be written in quadratic form.
8$z$ + 17$\sqrt{z}$ = 0 a. No b. Yes

#### Solution:

8z + 17z = 0

Recall that z = (z)2.

8z + 17z = 0 becomes, 8(z)2 + 17z = 0, a quadratic in z.

So, the given equation can be written in quadratic form.

6.
Verify if the equation can be written in quadratic form.
$x$4 + 4$x$2 = 5 a. No b. Yes

#### Solution:

x4 + 4x2 = 5

x4 + 4x2 - 5 = 0

(x2)2 + 4x2 - 5 = 0, a quadratic in x2.

So, the given equation can be written in quadratic form.

7.
Solve: ($d$ + 2)($d$ - 3) = 5$d$ + 10 a. - 2, 3 b. 8, - 2 c. - 8, - 2 d. - 8, 2

#### Solution:

(d + 2)(d - 3) = 5d + 10

d2 - d - 6 = 5d + 10
[Multiply.]

d2 - 6d - 16 = 0
[Group the like terms and simplify.]

(d - 8)(d + 2) = 0
[Factor.]

Therefore, d = 8, - 2

So, the solutions are 8, - 2.

8.
Solve: $n$ - 4$\sqrt{n}$ - 45 = 0 a. 81, 25$i$ b. 81, 25 c. 9, - 5 d. 81

#### Solution:

n - 4n - 45 = 0

(n)2 - 4n - 45 = 0

d2 - 4d - 45 = 0
[Replace n by d.]

(d - 9)(d + 5) = 0
[Factor.]

Therefore, d = 9, or d = - 5
[Solve for d.]

That is n = 9, or n = - 5
[Replace d by n.]

n = 9 gives, n = 81
[Square each side of the equation.]

Ignore n = - 5 as square root of a number is not negative.

So, the solution is 81.

9.
Solve: ($j$ + 8)2 - 11($j$ + 8) + 28 = 0 a. 0, -2 b. 0, 2 c. 1, 4 d. -1, -4

#### Solution:

(j + 8)2 - 11(j + 8) + 28 = 0

p2 - 11p + 28 = 0
[Replace (j + 8) by p.]

(p - 7)(p - 4) = 0
[Factor.]

Therefore, p = 7, or p = 4
[Solve for p.]

That is j + 8 = 7, or j + 8 = 4
[Replace p by (j + 8).]

j + 8 = 7 gives, j = -1 and j + 8 = 4 gives, j = -4
[Simplify.]

So, the solutions are -1 and -4.

10.
Verify if the equation can be written in quadratic form. $\frac{5}{s-5}+\frac{5}{5-s}$ = 4 a. Yes b. No

#### Solution:

5s-5 +55-s = 4

(s - 5)(5 - s)[5s-5 +55-s] = 4(s - 5)(5 - s)
[Multiply throughout by the LCD (s - 5)(5 - s).]

(25 - 5s) + (5s - 25) = 4(s - 5)(5 - s)

0 = 4 (- s2 + 10s - 25)

- 4s2 + 40s - 100 = 0, is a quadratic in s.

So, the given equation can be written in quadratic form.