Standard equation of a Quadratic Polynomial is ax2 + bx + c, and since the greatest degree of the polynomial is 2, it can have maximum of two roots. We will discuss all the possibilities of quadratic equation having two roots, one root or no root.
The graph of a quadratic polynomial depends on the coefficient of x2 i.e. ‘a’. If ‘a’ is positive, graph of quadratic polynomial will be a upright parabola (U-shaped) and if ‘a’ is negative, graph will be of the shape of inverted parabola. The logic is same as that of a linear polynomial as the coefficient of ‘x’ (y = mx + c) determines whether the line will be increasing or decreasing.
Let us plot a quadratic polynomial y = x2 – 3x – 4; While plotting the graph we should take care of few points. First of all, we should find out the roots of the polynomial as the roots will help us in determining the points where the graph will cut the X-axis. And secondly, we can also find out the y-intercept easily which will be constant part in the equation given. In this equation given, roots are (–1 &4) and y-intercept will be –4. Also, the graph of a quadratic polynomial will increase rapidly when we increase the value of x or decrease the value of x.
| X | –3 | –2 | –1 | 0 | 1 | 2 | 3 | 4 |
| Y | 14 | 6 | 0 | –4 | –6 | –6 | –4 | 0 |
On observing the table, x = –1 and x =4 are roots of the polynomial as value of y-coordinate is equal to 0. Also, y-intercept is –4 as the value of x-coordinate is equal to 0. And also as the value of ‘x’ starts increasing, value of ‘y’ increases rapidly. And when value of ‘x’ decreases, then also ‘y’ increases rapidly.
Let us plot one more graph of a quadratic polynomial with negative sign of ‘a’.
Y = –x2 – x + 6; Again we need to find out the roots of the polynomial, we need to equate the value of polynomial to zero.
So, we get –x2 – x + 6 = 0, which is same as x2 + x – 6 = 0. So, the roots are (2) & (–3) and the y-intercept is +6. (y = –x2 – x + 6).
After observing both the graphs, we would have got an idea that whenever the coefficient of x2 is positive, the graph will be a Upright-Parabola and when the coefficient of x2 is negative, the graph will be an Inverted-Parabola.
Factor Theorem: If we know roots of a quadratic expression, we can frame the factors and if we know the factors of a quadratic expression, we can get the roots of the equation. So, both factors and roots are interrelated.
For e.g. if a quadratic expression is given as y = x2 – 10x + 24. We can factorize the expression as y = (x – 6) (x – 4).
So, (x – 6) and (x – 4) are factors of the expression and to find the value of roots, we need to equate the expression to 0.
So, (x – 6)(x – 4) = 0 which gives the value of x as 4 or 6. These two are the roots. So, if we know the factors of an expression, we can equate it to 0 to get the roots. And if we know the roots, we can get the factors by subtracting the root from x.
For e.g., If its given that 2 and – 3 are the roots of a quadratic expression. So, we can frame the factors as (x – 2) and [x – (–3)]= (x + 3) are factors of the expression .
So, Factor Theorem states that if (x – α) is a factor of polynomial f(x), then on plug-in value of x = α in f(x), f(x) will be equal to 0.
E.g 1: If its given that (x –3) is a factor of 2x2 + 5x + k, find the value of k.
If (x –3) is given as the factor, then applying the factor theorem, we can say that x = 3 will be root of the given expression and on substituting x = 3 in expression, the value of expression must be equal to zero.
So, on plug-in 2 × 32 + 5 × 3 + k = 0; K is equal to –33.
We will learn some more applications of Factor theorem later on.
Sum of the roots and Product of the roots:
Roots of a quadratic expression ax2 + bx + c can be calculated in two ways:
i) By factorisation i.e. splitting the middle term.
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Now, the question arises if we want to find the sum of roots or product of roots of quadratic expression is it mandatory to find out the roots first?
Answer is no. We can directly find out sum of the roots and product of roots and use it as a formula. We will see the derivation now.
Let us assume the two roots of quadratic expression ax2 + bx + c as α and β.
Now using FACTOR THEOREM, if α and β are the roots then (x – α) and (x – β) will be the factors of the quadratic expression.
So, we can express as (x – α) × (x – β) = ax2 + bx + c.
On multiplying, we get [x2 – (α + β)x + αβ] = ax2 + bx + c.
The equation is not balanced as the coefficient of x2 on the L.H.S is 1, but the coefficient of x2 on R.H.S is ‘a’. So, to make it balanced we will divide the expression on the R.H.S by ‘a’.
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On observing the above equation, we can conclude one more point about quadratic expression.
A new quadratic equation can be expressed as x2 – (sum of the roots)x + product of roots = 0.
On equating the coefficient of ‘x’ and constant on both sides, we get
Now, we can use these two formulas directly in any quadratic expression. Let us see some applications.
E.g. 2: Construct a quadratic equation if it’s two roots are 4 and –5.
To frame quadratic equation, we need to find sum of the roots and product of the roots.
Sum of the roots = 4 + (–5) = –1.
And product of the roots = (4) × (–5) = –20.
So, the equation can be framed by substituting the values in x2 – (sum of the roots)x + product of roots = 0.
On plug-in values, we get x2 – (–1)x + (–20) = 0 or, x2 + x – 20 = 0.
E.g. 3: If p and q are the roots of equation x2 – 5x + 7 = 0, frame a new equation whose roots are (p + 5) and (q + 5).
To frame a new equation, we need to find the sum of roots and product of roots of the required equation.
Our required equation is x2 – (p + 5 + q + 5)x + (p + 5) (q + 5) = 0.
So, we need to know the value of (p + 5) + (q + 5) = p + q + 10. We already know the value of (p + q) from the first equation as p & q are roots of the first equation.
So, p + q = –b/a = –(–5)/1 = 5, then value of (p + q + 10) = 5 + 10 = 15.
Now, we should also know the value of the product of roots of the required equation which is (p + 5) × (q + 5).
So, (p + 5) × (q + 5) = pq + 5(p + q) + 25 = 7 + 5 × 5 + 25 = 57 [pq = product of roots of first equation whose value is c/a = 7].
So, now we can plug these values in equation x2 – (sum of the roots)x + product of roots = 0.
So, answer is x2 – 15x + 57 = 0.
E.g. 4: If (p + 3) and (q + 3) are roots of x2 + 8x – 10, then frame a equation whose roots are p & q.
The new equation with the roots p & q will be of the form x2 – (p + q)x + pq = 0.
That means we need to find the value of (p + q) and pq.
We know sum of roots of equation given in question i.e. (p + 3 + q + 3) = –8 which gives us the value of (p + q) = –14.
Similarly, we can find the product of roots from the equation given.
(p + 3) (q + 3) = –10, On multiplying we get [pq + 3(p + q) + 9] = –10.
On substituting the value of (p + q) obtained earlier, we can find the value of pq.
So, [pq –42 + 9] = –10, pq = 23
Now, we know the value of (p + q) = –14 and pq = 23.
So, the new equation framed is x2 + 14x + 23 =0.
E.g. 5: If two roots of equation 3x2 – 7x + 2k + 5 = 0 are reciprocal of each other, then find the value of k.
This is a pretty easy question which can be solved orally if we know that if the roots are reciprocal of each other leads to product of the roots being equal to 1.
For e.g. if one root is p, second root will be 1/p since roots are reciprocal and their product will always be 1.
We know the product of roots = c/a = (2k + 5)/3 which is equal to 1.
So, on solving k = –1.
E.g. 6: If two roots of equation 4x2 – (5p + 15)x – 4 = 0 are equal in magnitude but opposite in sign, find the value of p.
Again, a straightforward question if we know the logic that if the roots are equal in magnitude and opposite in sign, then the sum of the roots is always equal to zero.
If we assume one root as z, then the second root will be –z. And sum of those two roots will be 0.
We know sum of the roots given in the question is (5p + 15)/4 and equate it to 0.
We get (5p + 15)/4 = 0. On solving, we get value of p = –3.
Determinant:
Determinant is denoted by D whose value is equal to b2 – 4ac. It is called Determinant as it helps in determining the nature of roots of quadratic equation. We will take some examples which will make us understand the logic.
First case: if Determinant is positive, then the roots will be real & unequal.
Second case: If determinant is 0, then the roots will be real & equal. (Also pay attention the quadratic equation in 2nd case is a perfect square. So, we should keep in mind that determinant of a perfect square will always be zero which will give us real and equal roots).
Third case: If determinant is negative, then the roots will be imaginary as square root of a negative number is imaginary.
Now, we look at the positioning of the graphs in all these different cases.
E.g. 7: If the equation x2 – 15 – m(x – 4) = 0 has equal roots, find the value/s of m.
Before solving this problem, we need to rearrange the equation in standard form which is ax2 + bx + c = 0.
On rearranging, we get x2 – mx + (4m – 15) = 0.
Since, the equation has equal roots, determinant must be equal to zero.
Thus, on equating b2 – 4ac = 0, we get m2 – 4(4m – 15 ) = 0.
On solving, we get values of m as 6 or 10.
Common Roots: Sometimes two quadratic equations have one root in common or two roots in common. In such scenarios we will go back to basics of the definition of root. Roots are the point which intersects the X-axis or roots are the values which on substituting in quadratic expression, the value of expression becomes zero. So, if a pair of quadratic equation has one root in common, then on substituting the value of root in both the equations, value becomes zero.
For example, the expressions (x2 – 3x + 2) and (2x2 – 3x – 2) have (x – 2 ) as the common factor which means x = 2 is the common root of both the expressions. And if we substitute x = 2 in both the expressions, values of both the expressions will become zero.
E.g. 8: If (x + 3) is common factor of x2 – 5x + k = 0 and x2 – 6x – p = 0, then find the values of k and p.
This is a very easy problem based on common root. If (x + 3) is common factor to both the equations, then x = –3 will be the common root to both the equations and on substituting x =–3, the value of the equations will become zero.
So, plug-in value of x in both the equations and find out respective values of k & p.
In first equation, on substituting, we get 9 + 15 + k = 0; so, k = -24.
And in second equation, on substituting, we get 9 + 18 – p = 0; we get p = 27.
What if both the roots of two quadratic equations are common?
We can handle this situation in two ways:
First way: If both the roots of two equations are common, then we can equate the sum of roots of the first equation with sum of roots of second equation as both will be equal. Similarly, we can also equate product of roots of the first equation with the product of roots of the second equation as both will be equal.
Second way: If two equations are ax2 + bx + c =0 and dx2 + ex + f =0 having two roots in common, then the ratio of the coefficients of x2, x and constant will be equal.
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For e.g. (x2 + 3x + 2 = 0) and (2x2 + 6x + 4= 0) are two equations with both the roots common and the ratio just discussed earlier will be same.
Cubic Equation:
Cubic equation can have maximum of 3 roots and as we saw there is relationship between roots of a quadratic equation, similarly there exists certain relationship among roots of a cubic equation.
If p, q and r are the roots of cubic equation ax3 + bx2 + cx + d =0.
Then sum of roots taken one at time i.e. (p + q + r) = –b/a.
Sum of roots taken 2 at a time i.e. (pq + pr + qr) = c/a.
And sum of roots taken 3 at a time or product of roots i.e. (pqr) = –d/a.
Also, we can notice the pattern that the negative & positive sign changes alternately.
E.g. 9: If p, q and r are the roots of cubic equation 2x3 – 4x2 + 6x –10 = 0, then find value of p3 + q3 + r3.
Before doing anything, we should see what is the requirement to find out value of p3 + q3 + r3?
We know, p3 + q3 + r3 = (p + q+ r)[p2 + q2 + r2 – (pq + pr + qr)] + 3pqr.
We know the value of (p + q + r) which is sum of roots of cubic equation given equal to 2. Similarly we also know the value of (pq + pr + qr) which is sum of roots taken 2 at a time equal to 3. And finally, we also know value of pqr = 5.
Now, to find out value of p3 + q3 + r3, we should also know the value of p2 + q2 + r2.
No, the value of p2 + q2 + r2 can be found out with the help of an identity.
So, p2 + q2 + r2 = [p + q + r]2 – 2(pq + pr + qr] = 4 – 6 = –2.
Now, we can plug-in values of all of them in formula of p3 + q3 + r3 to obtain its value.
So, p3 + q3 + r3 = (2) [– 2 – (3)] + 3 × 5 = –10 + 15 = 5.
So, this problem is an example of application of identities and relations between the roots of a cubic equation. If we are comfortable with these formulas, we can solve these problems orally.
Exercise:
1. If p and q are the roots of x2 – cx + b = 0, find the values of
a) p2 + q2.
b) p3 + q3.
2. If p and q are the roots of equation x2 – 5x + 7 =0, then frame a new equation whose roots are p/q and q/p.
3. For what values of ‘p’ will the equation x2 – 2x(1 + 3p) + 7(3 + 2p) = 0 have real & equal roots?
4. For what value of ‘k’ will the equation x2 – 5k(x + 2) – 2(x + 3) = 0 have roots equal in magnitude but opposite in sign?
5. If p and q are the roots of equation 2x2 – 6x + 3 = 0, then frame a equation whose roots are (p – q)2 and (p + q)2.
6. If p, q and r are the roots of the cubic equation 5x3 – 3x2 + 4x –3 =0, find the value of
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7. If the difference between the roots of the equation 2x2 – mx + 15 = 0 is 1.5, find m.
8. If one of the root of the equation x2 – 2x – 15 = 0 is the same as one root of the equation x2 – 10x + p =0, find p.
10. If the equation 2x2 – px + 8 = 0 has imaginary roots, then how many positive integral values can ‘p’ take?
11. Which is the least positive integral value that ‘p’ can assume if the roots of the equation 4x2 – px + 5 = 0 are real and unequal?
12. How many integral values of ‘p’ satisfy the relation x2 – 5px + 4p2 + 1> 0 for all real values of x?
13. If both the roots of the equation x2 – 6px + 2 – 2p =0 exceed 5, then find the range of ‘p’.
14. Find the range of ‘p’ if the roots of the equation 2x2 – 5x + p2 – 6p + 8 = 0 are opposite in sign?
15. What is the condition for one root of the quadratic equation px2 + qx + r =0 to be thrice the other? (Answer in terms of p, q and r).
16. While solving a quadratic equation, Ram made a mistake in copying the constant term and got the roots as 1 and –5. While Shyam made a mistake in copying the coefficient of x and got the roots as –3 and 4. Find the correct equation and its roots.
17. If one of the roots of the equation x2 + px – 28 = 0 is –4 and roots of the equation x2 + px + r = 0 are real and equal, then find the value of r.
18. If two quadratic equations px2 + x + 5 = 0 and 2x2 + x + q = 0 have a common root x = 2, then which of the following statements hold true?
a) 4p + q = –17 b) p + q = –8.25 c) p – q = 11.75 d) 4p – q = –3
Answer:
1.) (a) c2 – 2b 1. (b) c3 – 3bc.
2. 7x2 – 11x + 7 = 0. 3. 2 or –10/9 4. – 0. 4 5. X2 – 12x + 27 = 0
6. –31/15 7. ±√129 8. 25 or –39 9. None
10. 7 positive integral values 11. 9 12. 1 13. P > 1.66 or P < –11.5.
14. 2 < P < 4 15. 3q2 – 16pr = 0 16. The correct equation is x2 + 4x – 12 = 0, and roots are 2, –6. 17. r = 2.25 18. (a)