Solving quadratic inequalities worksheet for grade 11 :
Here we are going to see some practice questions on solving quadratic inequalities.
Here we shall learn to solve the quadratic inequalities ax^{2} + bx + c < 0 or ax^{2} + bx + c > 0.
(i) First solve ax^{2} + bx + c = 0.
(ii) If there are no real solutions, then one of the above inequality holds for all x ∈ R
(iii) If there are real solutions, which are called critical points, then label those points on the number line.
(iv) Note that these critical points divide the number line into disjoint intervals. (It is possible that there may be only one critical point.)
(v) Choose one representative number from each interval.
(vi) Substitute that these representative numbers in the inequality.
(vii) Identify the intervals where the inequality is satisfied.
Problem 1 :
Solve :
2x^{2} + x - 15 ≤ 0
Problem 2 :
Solve :
-x^{2} + 3x - 2 ≥ 0
Problem 1 :
Solve :
2x^{2} + x - 15 ≤ 0
Solution :
First let us solve the given quadratic equation by factoring.
2x^{2} + x − 15 = 0
2x^{2} + 6x - 5x − 15 = 0
2x(x + 3) - 5(x + 3) = 0
(2x - 5) (x + 3) = 0
2x - 5 = 0 x + 3 = 0
2x = 5 x = -3
x = 5/2
The critical numbers are -3 and 5/2.
Now let us mark them in number line.
Applying any values within the interval, we get
Intervals |
Signs of factors (2x-5)(x+3) |
Sign of given inequality |
(-∞, -3] say x = - 4 |
(-) (-) |
+ |
[-3, 5/2] say x = 0 |
(-) (+) |
- |
[5/2, ∞) say x = 3 |
(+) (+) |
+ |
From the above table, we come to know that the interval [-3, 5/2] satisfies the given inequality.
Hence, the solution is [-3, 5/2].
Problem 2 :
Solve :
-x^{2} + 3x - 2 ≥ 0
Solution :
First let us solve the given quadratic equation by factoring.
The coefficient of x must be positive, so we have to multiply the inequality by negative.
x^{2} - 3x + 2 ≤ 0
Multiply the equation by negative.
x^{2} - 3x + 2 = 0
x^{2} - 1x - 2x + 2 = 0
x (x - 1) - 2(x - 1) = 0
(x - 2) (x - 1) = 0
x - 2 = 0 x - 1 = 0
x = 2 x = 1
Writing them as intervals, we get
(-∞, 1] [1, 2] [2, ∞)
Applying any values within the interval, we get
Intervals |
Signs of factors (x-2)(x-1) |
Sign of given inequality |
(-∞, 1] say x = 0 |
(-) (-) |
+ |
[1, 2] say x = 1.5 |
(-) (+) |
- |
[2, ∞) say x = 3 |
(+) (+) |
+ |
From the above table, we come to know that the interval [1, 2] satisfies the given inequality.
Hence, the solution is [1, 2].
After having gone through the stuff given above, we hope that the students would have understood, how to solve quadratic inequalities.
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