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CAT 2025 Lesson : Quadratic Equations - Maxima, Minima & Descrates Rule

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6. Descartes' rule

Where f(x)f(x) is a polynomial and the terms are arranged in descending order of their powers,
1)1) Maximum number of positive real roots of f(x)f(x) equals the number of sign changes in coefficients of subsequent terms of f(x).
22) Maximum number of negative real roots of f(x) equals the number of sign changes in coefficients of subsequent terms of f(-x).
33) The actual number of positive or negative real roots can differ from their respective maximum values by a multiple of 2.

For instance, if the number of sign changes of f(x)f(x) is 77, then number of positive real roots can be 7,5,37, 5, 3 or 11. Likewise, if the number of sign changes of f(−x)f(-x) is 66, then number of negative real roots can be 6,4,26, 4, 2 or 00.

Example 26

Which of the following can be the number of negative real roots if x7+ax5−bx4+cx2+x+5=0x^{7} + ax^{5} - bx^{4} + cx^{2} + x + 5 = 0, where a,ba, b and cc are positive real numbers?

(1) 00            (2) 11            (3) 22            (4) 55           

Solution

Where f(x)=x7+ax5−bx4+cx2+x+5=0f(x) = x^{7} + ax^{5} - bx^{4} + cx^{2} + x + 5 = 0,
f(−x)=−x7−ax5−bx4+cx2−x+5f(-x) = -x^{7} - ax^{5} - bx^{4} + cx^{2} - x + 5

As a,ba, b and cc are positive, the sign changes in the term are − − − + − +-\space -\space -\space +\space -\space +

The sign has changed 33 times, i.e., in the x2,xx^{2}, x and constant terms.

∴ Maximum number of negative roots is 33.

Number of negative real roots will be equal to or lower than 33 by a multiple of 22.

Possible number of negative real roots are 33 and 11. Only the number 11 is there in the answer options.

Answer: (2) 1(2) \space 1


7. Maxima and Minima of Quadratic Expression
A quadratic expression f(x)=ax2+bx+cf(x) = ax^{2} + bx + c, where a>0a \gt 0 forms a U-shaped curve.

Here, the minimum value of f(x)f(x) is at x=−b2ax = \dfrac{-b}{2a}

And, the minimum value of f(x)=(4ac−b2)4af(x) = \dfrac{(4ac - b^{2})}{4a}
A quadratic expression f(x)=ax2+bx+cf(x) = ax^{2} + bx + c, where a<0a \lt 0 forms an inverted U-shaped curve

Here, the maximum value of f(x)f(x) is at x=−b2ax = \dfrac{-b}{2a}

And, the maximum value of f(x)=(4ac−b2)4af(x) = \dfrac{(4ac - b^{2})}{4a}


Note that (4ac−b2)4a\dfrac{(4ac - b^{2})}{4a} is the expression we get by substituting x=−b2ax = \dfrac{-b}{2a} in f(x)f(x).

Example 27

What is the maximum value of f(x)=−2x2−5x+6f(x) = -2x^{2} - 5x + 6?

Solution

Maximum value of f(x)=(4ac−b2)4a=(4×−2×6−(−5)2)4×−2=−48−25−8=738f(x) = \dfrac{(4ac - b^{2})}{4a} = \dfrac{(4 \times -2 \times 6 - (-5)^{2})}{4 \times -2} = \dfrac{-48 - 25}{-8} = \dfrac{73}{8}

Answer: 738\dfrac{73}{8}


Example 28

At what value of xx, does f(x)=x2−5x+6f(x) = x^{2} - 5x + 6 attain minimum?

Solution

As a=1a = 1 is positive, this function will attain minimum.

f(x)f(x) will attain minimum at x=−b2a=52x = \dfrac{-b}{2a} = \dfrac{5}{2}

Answer: 52\dfrac{5}{2}


Example 29

If α\alpha and β\beta are the roots of the equation x2−3mx+m−5=0x^{2} - 3mx + m - 5 = 0, then what is the minimum value of α2+β2+5αβ\alpha^{2} + \beta^{2} + 5 \alpha \beta?

Solution

Sum of roots =α+β=3m= \alpha + \beta = 3m
Product of roots =αβ=m−5= \alpha \beta = m - 5

α2+β2+5αβ\alpha^{2} + \beta^{2} + 5 \alpha \beta
⇒ α2+β2+2αβ+3αβ \alpha^{2} + \beta^{2} + 2 \alpha \beta + 3 \alpha \beta
⇒ (α+β)2+3αβ (\alpha + \beta)^{2} + 3 \alpha \beta
⇒ 9m2+3m−15 9m^{2} + 3m - 15

Minimum value of this expression =(4ac−b2)4a= \dfrac{(4ac - b^{2})}{4a} =((4×9×−15)−32)4×9= \dfrac{((4 \times 9 \times -15) -3^{2})}{4 \times 9} =−614= \dfrac{-61}{4}

Answer: −614\dfrac{-61}{4}


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