Equations quadratic in form

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Algebra 3

EQUATIONS QUADRATIC IN FORM
The equation x4 + x2 – 12 = 0 is not a quadratic in x, but it is a quadratic in x2 . That is, if u = x2,
we get u2 + u –12 = 0, a quadratic equation. This equation can be solved for u and, in turn, by using u = x2, we can find the solutions x for the original equation.
In general, if an appropriate substitution utransforms an equation into one of the form
au2 + bu + c = 0
then the original equation is called an equation of the quadratic type or an equation quadratic in form. After you are told an equationis quadratic in form, it is easy enough to see it, but some practice is needed to enable you to recognize them on your own.
Solving Equations That Are Quadratic in Form
Find the real solutions ofthe equation: (x + 20)2 + 11(x + 2) – 12 = 0
Solution: For this equation, let u = u + 2. Then u2 = (x + 2)2, and the original equation,
(x + 2)2 + 11(x + 2) – 12 = 0
Becomes u2 + 11u -12 = 0 Let u= x +2
(u + 12) (u – 1) m= 0 Factor
u = -12 or u = 1 Solve
But we want to solve for x. Because u = x + 2, we have
x + 2 = -12 or x + 2 = 1
x = -14 x = -1
Check: x= -14: (-14 + 2)2 + 11(-14 + 2) -12
= (-12)2 + 11(-12) – 12 = 144 – 132 – 12 = 0
X = -1: (-1 + 2)2 + 11(-1 + 2) – 12 = 1 +11 -12 = 0
The original equation has the solution set { -14 , -1}For the following examples, find the real solutions of each equation.
1) x4 – 5x2 + 4 =0 2) 3x4 – 2x2 – 1 = 0 3) x6 + 7x3 – 8 = 0

4) (x + 2)2 + 7(x + 2) + 12 = 0 5) (3x + 4)2 -6(3x + 4) +9 = 0 6) 2(s + 1)2 – 5(s + 10) = 3

7) x – 4x = 0 8) x + = 20 9) 4x1/2 - 9x1/4 + 4 = 0

10) t1/2 - 2t1/4 + 2 = 0 11) = x 12) x2 + 3x + = 6
13) 2 =+ 2 14) 3x-2 – 7x-1 – 6 = 0
15) 2x2/3 – 5x1/3 -3 = 0 16) = 9
Answers: 1) {-1,-1, 1, 2} 2) {-1, 1} 3) {-2, 1} 4) {-6, -5} 5) {- } 6) {- , 2}
7) {0, } 8) { 16 } 9) { 1 } 10)...