Question

how are quadratic inequalities different from linear inequalities​

Answer 1

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ᴀ ʟɪɴᴇᴀʀ ᴇǫᴜᴀᴛɪᴏɴ ɪs ᴏɴᴇ ᴡʜᴇʀᴇ ᴛʜᴇ sᴏʟᴜᴛɪᴏɴ sᴇᴛ ᴄᴀɴ ʙᴇ ᴅʀᴀᴡɴ ᴀs ᴀ ʟɪɴᴇ. ғᴏʀ ᴇxᴀᴍᴘʟᴇ, sᴏᴍᴇᴛʜɪɴɢ ᴏғ ᴛʜᴇ ғᴏʀᴍ ʏ = ᴍx + ʙ. ᴛʜᴇ ɪᴍᴘᴏʀᴛᴀɴᴛ ᴅɪғғᴇʀᴇɴᴄᴇ ɪɴ ᴛʜᴇ ᴇǫᴜᴀᴛɪᴏɴ ɪs ᴛʜᴀᴛ ʙᴏᴛʜ x ᴀɴᴅ ʏ ᴀʀᴇ ɴᴏᴛ ʀᴀɪsᴇᴅ ᴛᴏ ᴀɴʏ ᴘᴏᴡᴇʀs. ᴀ ǫᴜᴀᴅʀᴀᴛɪᴄ ᴇǫᴜᴀᴛɪᴏɴ ɪs ᴏɴᴇ ᴡʜᴇʀᴇ ᴛʜᴇ sᴏʟᴜᴛɪᴏɴ sᴇᴛ ᴄᴀɴ ʙᴇ ᴅʀᴀᴡɴ ᴀs ᴀ ᴘᴀʀᴀʙᴏʟᴀ ᴡʜɪᴄʜ ɪs ᴀ sᴘᴇᴄɪᴀʟ ᴄᴜʀᴠᴇ sʜᴀᴘᴇ.❤

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Answer 2

Step-by-step explanation:

Linear Inequalities Linear inequalities are those inequalities in which the highest power of the variable is 1. These are the simple inequalities to solve. It requires the knowledge of some basic algebraic rules to solve them. Let us take an example.

Example 1: Solve the inequality; (3x+2/-3)< Solution : We have (3x+2/-3)< => 3x + 2 > - 15 (The sign of the inequality has been changed as it is multiplied by - 3) 3x > - 17 => x >(-17/3) is the solution of the given inequality

Quadratic inequalities are those inequalities in which the highest power of the variable is 2.

Now to solve such inequalities use the following steps:

Make right hand side of the inequality equal to zero by transposing the terms (if any) from left hand side.

Factorize the expression on the left hand side and make sure that the coefficient of 'x' is positive in each factor.

Equate to zero both the factors to get the critical points. Plot these points on the number line. There will be three regions on the number line and the rightmost region will give you positive inequalities, middle one will give you negative inequalities and the leftmost part will again give you positive inequalities. You will understand this from the following example.

Example 2: Solve the inequality x2 + x - 28 < 2.

Solution : We have x2 + x - 28 < 2

Step I: Make the right hand side equal to zero, we get x2 + x - 30 < 0 Step II: Factorize the equation, we get x2 +x - 30 < 0 => x2 + 6x - 5x - 30 < 0 => x(x + 6) - 5(x + 6) < 0 => (x - 5) (x + 6) < 0 Step III: Equate to zero both the factors to get the critical points, we get x = 5, - 6 Plot these points on the number line, we get  Now as our inequality is a negative inequality, so the middle part is our solution. Hence the solution is (- 6, 5).

 

Note: If the original inequality is x2 +x - 30 > 0, then the solution will follow the same above steps and the solution of the inequality will be (- ∞, - 6) U (5,∞) i.e. all the values for which the graph is showing the positive values.