Equations Reducible to Quadratic Equations

2. Equations of Higher Degree 3. Equations with Square Root Signs 5. Logarithmic Equations 1. 4. Exponential Equations They include the following types. Fractional Equations Some equations that are not in the quadratic form can be reduced to quadratic equations.

Fractional Equations 5 6 = – x 2 3 1 + e.g. and The two given equations are fractional equations. They can be solved by first reducing them to quadratic equations.

Reduce the fractional equation to a quadratic equation. Solve . 5 6 = – x Step 1 Reduce the fractional equation to a quadratic equation. 5 6 = – x  Note that x ≠ 0. 6 – x2 = 5x  Multiply both sides by x. 6 5 2 = – + x

Solve . 5 6 = – x Step 2 Solve the quadratic equation obtained. Remember to check whether the answers obtained satisfy the original equation.

2 3 1 = + – x Solve .  Note that x ≠ 0 and x ≠ 1.  Multiply both sides by the L.C.M. of x – 1 and x, i.e. x(x – 1).

Follow-up question Solve . | | 2 ) 1 ( 3 - = ø ö ç è æ x 2 ) 1 ( 3 - =  Note that x ≠ 0. 2 3 - = + x  Expand the expression on the L.H.S. 3 7 2 = + - x 3 7 2 = + - x  Multiply both sides by x. ) 3 )( 1 2 ( = - x 3 or 2 1 = x ∴  Check whether the answers obtained satisfy the original equation.

Equations of Higher Degree Some equations of higher degree can be solved by reducing them to quadratic equations, for example x4 – 6x2 – 27 = 0 and 8x6 – 7x3 – 1 = 0.

Solve ax2n + bxn + c = 0, where a 0 and n is a positive number. Solve x4 – 6x2 – 27 = 0. a = 1 n = 2 Step 1 Substitute xn = u into the equation to obtain quadratic equation in u, au2 + bu + c = 0. By substituting x2 = u into the equation x4 – 6x2 – 27 = 0, we have u2 u u x4 = u2

Solve ax2n + bxn + c = 0, where a 0 and n is a positive number. Solve x4 – 6x2 – 27 = 0. Step 2 Solve au2 + bu + c = 0 to find the value(s) of u. u2 – 6u – 27 = 0 (u – 9)(u + 3) = 0 u = 9 or u = –3 Step 3 Since x2 = u, we have x2 = 9 or x2 = –3 (rejected) Solve xn = u to find the value(s) of x. x = ±3 ∴ x = ±3

We must not stop here since we are solving for x, not u. Solve 8x6 – 7x3 – 1 = 0. By substitute x3 = u into the equation 8x6 – 7x3 – 1 = 0, we have  x6 = (x3)2 = u2 (8u + 1)(u – 1) = 0 8 1 – = u or u = 1 Since x3 = u, we have 8 1 – = x3 or x3 = 1 We must not stop here since we are solving for x, not u. 2 1 – = x or x = 1 2 1 – = x or x = 1 ∴

Follow-up question Solve x5 – 17x3 + 16x = 0. Remember not to cancel the factor x, otherwise we will miss one of the solutions, x = 0. x5 – 17x3 + 16x = 0 x(x4 – 17x2 + 16) = 0 x = 0 or x4 – 17x2 + 16 = 0 ......(1) By substituting x2 = u into (1), we have u2 – 17u + 16 = 0 (u – 1)(u – 16) = 0 u = 1 or u = 16 Since x2 = u, we have x2 = 1 or x2 = 16 x = ±1 or x = ±4 x = 0

Equations with square root signs Let me show you how to solve . 1 5 = + - x

Group the terms without square root signs on the same side. Solve . 1 5 = + - x Step 1 Group the terms without square root signs on the same side. 1 5 = + - x 5 1 + = - x Step 2 Square both sides to eliminate the square root sign. ( ) 1 = - x 5 + 2 2 5 1 2 + = - x 4 3 2 = - x

Step 3 Solve the quadratic equation obtained. Squaring both sides of an equation in step 2 may create unwanted roots. Hence, we need to do the following step.

Check whether the results satisfy the original equation. Step 4 Check whether the results satisfy the original equation. Checking: When x = –1, 5 = + - x 1 = - 3 1 ¹ Hence, –1 is not a solution of the equation . 1 5 = + - x When x = 4, 5 = + - x 4 = 1 ∴ The solution of the equation is x = 4.

Follow-up question Solve . 1 2 = - x 1 2 = - x 1 2 = - x ) ( 1 2 = - x  Group the terms without square root signs on one side. ) ( 1 2 = - x  Square both sides. 1 4 2 = + - x 1 5 4 2 = + - x ) 1 )( 4 ( = - x 1 or 4 = x

∴ The solution of the equation is x = 1. Checking: When , 4 1 = x 4 1 2 - | ø ö ç è æ = x = 1 ¹ Hence, is not a solution of the equation . 4 1 2 = - x When x = 1, 1 ) ( 2 - = x 1 = ∴ The solution of the equation is x = 1.

Alternative Solution By substituting into the equation , we have 1 2 = - x u 2u2 – u = 1  2 ) ( u x = 2u2 – u – 1 = 0 (u – 1)(2u + 1) = 0 2 1 – = u u = 1 or Since , we have u x = 2 1 – = x or = 1 (rejected)  is always non-negative. x ∴ x = 1

Both exponential equations and logarithmic equations can sometimes be reduced to quadratic equations by substitution, and be solved.

Exponential Equations Let look at an example of solving the exponential equation 22x + 2(2x) – 3 = 0 first.

Solve 22x + 2(2x) – 3 = 0. 22x + 2(2x) – 3 = 0 (2x)2 + 2(2x) – 3 = 0 ......(1) By substitute 2x = u into (1), we have u2 + 2u – 3 = 0 (u – 1)(u + 3) = 0 u = 1 or u = –3 Since 2x = u, we have 2x = 1 or 2x = –3 (rejected)  2x is always positive. 2x = 20 ∴ x = 0

Follow-up question Solve 22x – 5(2x+1) + 16 = 0. By substituting 2x = u into (1), we have u2 – 10u + 16 = 0 (u – 2)(u – 8) = 0 u = 2 or u = 8

Since 2x = u, we have 2x = 2 or 2x = 8 2x = 21 or 2x = 23 x = 1 or x = 3 ∴

Logarithmic Equations Now, let us solve the logarithmic equations (log x)2 + 2 log x – 3 = 0.

Solve (log x)2 + 2 log x – 3 = 0. By substitute log x = u into the equation (log x)2 + 2 log x – 3 = 0, we have Since log x = u, we have log x = 1 or log x = –3 x = 101 or x = 10–3 1000 1 x = 10 or x = ∴

Follow-up question Solve (log x)2 – log x2 – 8 = 0. By substitute log x = u into (1), we have u2 – 2u – 8 = 0 (u – 4)(u + 2) = 0 u = 4 or u = –2

Since log x = u, we have log x = 4 or log x = –2 x = 104 or x = 10–2 100 1 x = 10 000 or x = ∴