Quadratic Equations — Exercise 5.4

Find the discriminant of the following quadratic equations and hence find the nature of roots :

(i) 3x² - 5x - 2 = 0

(ii) 2x² - 3x + 5 = 0

(iii) 16x² - 40x + 25 = 0

(iv) 2x² + 15x + 30 = 0

Answer

(i) The given equation is 3x² - 5x - 2 = 0.

Comparing it with ax² + bx + c = 0, we get
a = 3 , b = -5 , c = -2

∴ Discriminant = b² - 4ac

Putting values of a, b, c in formula

$(-5)^2 - 4 \times 3 \times -2 \\[0.5em] = 25 + 24 \\[0.5em] = 49 \gt 0$

Discriminant = 49; Hence, the given equation has two distinct real roots.

(ii) The given equation is 2x² - 3x + 5 = 0.

Comparing it with ax² + bx + c = 0, we get
a = 2 , b = -3 , c = 5
∴ Discriminant = b² - 4ac
Putting values of a, b, c in formula

$(-3)^2 - 4 \times 2 \times 5 \\[0.5em] = 9 - 40 \\[0.5em] = -31 \lt 0$

Discriminant = -31; Hence, the given equation has no real roots.

(iii) The given equation is 16x² - 40x + 25 = 0.
Comparing it with ax² + bx + c = 0, we get
a = 16 , b = -40 , c = 25
∴ Discriminant = b² - 4ac
Putting values of a, b, c in formula

$(-40)^2 - 4 \times 16 \times 25 \\[0.5em] = 1600 - 1600 \\[0.5em] = 0$

Discriminant = 0; Hence, the given equation has two equal real roots.

(iv) The given equation is 2x² + 15x + 30 = 0.
Comparing it with ax² + bx + c = 0, we get
a = 2 , b = 15 , c = 30
∴ Discriminant = b² - 4ac
Putting values of a, b, c in formula

$(15)^2 - 4 \times 2 \times 30 \\[0.5em] = 225 - 240 \\[0.5em] = -15 \lt 0$

Discriminant = -15; Hence, the given equation has no real roots.

Discuss the nature of the roots of the following quadratic equations :

(i) $3x^2 - 4\sqrt{3}x + 4 = 0$

(ii) $x^2 - \dfrac{1}{2}x + 4 = 0$

(iii) $-2x^2 + x + 1 = 0$

(iv) $2\sqrt{3}x^2 - 5x + \sqrt{3} = 0$

Answer

(i) The given equation is $3x^2 - 4\sqrt{3}x + 4 = 0$.

Comparing it with ax² + bx + c = 0, we get
a = 3 , b = $-4\sqrt{3}$ , c = 4
∴ Discriminant = b² - 4ac
Putting values of a, b, c in formula

$(-4\sqrt{3})^2 - 4 \times 3 \times 4 \\[0.5em] = 48 - 48 \\[0.5em] = 0$

Hence the given equation has two equal real roots.

(ii) The given equation is $x^2 - \dfrac{1}{2}x + 4 = 0$.
Comparing it with ax² + bx + c = 0, we get
a = 1 , b = $-\dfrac{1}{2}$ , c = 4
∴ Discriminant = b² - 4ac
Putting values of a, b, c in formula

$\Big(-\dfrac{1}{2}\Big)^2 - 4 \times 1 \times 4 \\[1em] = \dfrac{1}{4} - 16 \\[1em] = \dfrac{1 - 64}{4} \text{ Taking L.C.M. } \\[1em] = -\dfrac{63}{4} \lt 0$

Hence the given equation has no real roots.

(iii)The given equation is -2x² + x + 1 = 0.
Comparing it with ax² + bx + c = 0, we get
a = -2 , b = 1 , c = 1
∴ Discriminant = b² - 4ac
Putting values of a, b, c in formula

$(1)^2 - 4 \times -2 \times 1 \\[0.5em] = 1 + 8 \\[0.5em] = 9 \gt 0 \\[0.5em]$

Hence, the given equation has two distinct real roots.

(iv) The given equation is $2\sqrt{3}x^2 - 5x + \sqrt{3} = 0$.
Comparing it with ax² + bx + c = 0, we get
a = $2\sqrt{3}$ , b = -5 , c = $\sqrt{3}$
∴ Discriminant = b² - 4ac
Putting values of a, b, c in formula

$(-5)^2 - 4 \times 2\sqrt{3} \times \sqrt{3} \\[0.5em] = 25 - 24 \\[0.5em] = 1 \gt 0$

Hence, the given equation has two distinct real roots.

Find the nature of roots of the following quadratic equations :

(i) $x^2 - \dfrac{1}{2}x - \dfrac{1}{2} = 0$

(ii)$x^2 - 2\sqrt{3}x - 1 = 0$

If real roots exist, find them.

Answer

(i) The given equation is $x^2 - \dfrac{1}{2}x - \dfrac{1}{2} = 0$

Comparing it with ax² + bx + c = 0, we get
a = 1 , b = $-\dfrac{1}{2}$ , c = $-\dfrac{1}{2}$

∴ Discriminant = b² - 4ac

Putting values of a, b, c in formula

$\Big(-\dfrac{1}{2}\Big)^2 - 4 \times 1 \times -\dfrac{1}{2} \\[1em] = \dfrac{1}{4} + 2 \\[1em] = \dfrac{1 + 8}{4} \\[1em] = \dfrac{9}{4} \\[1em]$

Discriminant = $\dfrac{9}{4}$
Since Discriminant > 0, hence the given equation has two distinct real roots.

The roots of the equation are given by:

$x = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2a} \\[1em] \Rightarrow x = \dfrac{-\Big(-\dfrac{1}{2}\Big) ± \sqrt{\Big(-\dfrac{1}{2}\Big)^2 - 4\times 1 \times -\dfrac{1}{2}}}{2 \times 1} \\[1em] \Rightarrow x = \dfrac{\dfrac{1}{2} ± \sqrt{\dfrac{1}{4} + 2}}{2} \\[1em] \Rightarrow x = \dfrac{\dfrac{1}{2} ± \sqrt{\dfrac{9}{4}}}{2} \\[1em] \Rightarrow x = \dfrac{\dfrac{1}{2} + \sqrt{\dfrac{9}{4}}}{2} \text{ or } \dfrac{\dfrac{1}{2} - \sqrt{\dfrac{9}{4}}}{2} \\[1em] \Rightarrow x = \dfrac{\dfrac{1}{2} + \dfrac{3}{2}}{2} \text{ or } \dfrac{\dfrac{1}{2} - \dfrac{3}{2}}{2} \\[1em] \Rightarrow x = \dfrac{\dfrac{4}{2}}{2} \text{ or } \dfrac{-\dfrac{2}{2}}{2} \\[1em] \Rightarrow x = \dfrac{4}{4} \text{ or } -\dfrac{2}{4} \\[1em] \Rightarrow x = 1 \text{ or } -\dfrac{1}{2}$

Hence roots of the given equations are 1 , $-\dfrac{1}{2}$.

(ii) The given equation is $x^2 - 2\sqrt{3}x - 1 = 0$
Comparing it with ax² + bx + c = 0, we get
a = 1 , b = $-2\sqrt{3}$ , c = -1
∴ Discriminant = b² - 4ac
Putting values of a, b, c in formula

$(-2\sqrt{3})^2 - 4 \times 1 \times -1 \\[1em] = 12 + 4 \\[1em] = 16 \\[1em]$

Discriminant = 16 ,
Since Discriminant > 0, hence given equation have real and distinct roots.

The roots of the equation are given by

$x = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2a} \\[1em] \Rightarrow x = \dfrac{-(-2\sqrt{3}) ± \sqrt{(-2\sqrt{3})^2 - 4\times 1 \times -1}}{2 \times 1} \\[1em] \Rightarrow x = \dfrac{2\sqrt{3} ± \sqrt{12 + 4}}{2} \\[1em] \Rightarrow x = \dfrac{2\sqrt{3} ± \sqrt{16}}{2} \\[1em] \Rightarrow x = \dfrac{2\sqrt{3} + 4}{2} \text{ or } \dfrac{2\sqrt{3} - 4}{2} \\[1em] \Rightarrow x = \sqrt{3} + 2 \text{ or } \sqrt{3} - 2 \\[1em]$

Hence roots of the given equations are , $\sqrt{3} + 2 , \sqrt{3} - 2$ .

Without solving the following quadratic equations, find the value of 'p' for which the given equations have real and equal roots :

(i) px² - 4x + 3 = 0

(ii) x² + (p - 3)x + p = 0

Answer

(i) The given equation is px² - 4x + 3 = 0

Comparing it with ax² + bx + c = 0, we get
a = p , b = -4 , c = 3

$\therefore \text{Discriminant} = b^2 - 4ac \\[1em] = (-4)^2 - 4 \times p \times 3 \\[1em] = 16 - 12p$

For equal roots, discriminant = 0

$\Rightarrow 16 - 12p = 0 \\[1em] \Rightarrow 16 = 12p \\[1em] \Rightarrow p = \dfrac{16}{12} \\[1em] p = \dfrac{4}{3}$

Hence the value of p is $\dfrac{4}{3}$ .

(ii) The given equation is x² + (p - 3)x + p = 0.

Comparing with ax² + bx + c = we obtain,
a = 1 , b = (p - 3) , c = p

$\therefore \text{Discriminant} = b^2 - 4ac \\[1em] = (p - 3)^2 - 4 \times 1 \times p \\[1em] = p^2 + 9 - 6p - 4p \\[1em] = p^2 + 9 - 10p$

For equal roots, discriminant = 0

$\Rightarrow p^2 + 9 - 10p = 0 \\[1em] \Rightarrow p^2 - 10p + 9 = 0 \\[1em] \Rightarrow p^2 - 9p - p + 9 = 0 \\[1em] \Rightarrow p(p - 9) - 1(p - 9) = 0 \\[1em] \Rightarrow (p - 1)(p - 9) = 0 \\[1em] \Rightarrow (p - 1) = 0 \text{ or } p - 9 = 0 \\[1em] \Rightarrow p = 1 \text{ or } p = 9 .$

Hence the value of p is 1, 9.

Find the values of k for which each of the following quadratic equation has equal roots :

(i) x² + 4kx + (k² - k + 2) = 0

(ii) (k - 4)x² + 2(k - 4)x + 4 = 0

Answer

(i) The given equation is x² + 4kx + (k² - k + 2) = 0.

Comparing with ax² + bx + c = we obtain,
a = 1 , b = 4k , c = (k² - k + 2)

$\therefore \text{Discriminant} = b^2 - 4ac \\[1em] = (4k)^2 - 4 \times 1 \times (k^2 - k + 2) \\[1em] = 16k^2 - 4(k^2 - k + 2) \\[1em] = 16k^2 - 4k^2 + 4k - 8 \\[1em] = 12k^2 + 4k - 8$

For equal roots, discriminant = 0

$\Rightarrow 12k^2 + 4k - 8 = 0 \\[1em] \Rightarrow 12k^2 + 12k - 8k - 8 = 0 \\[1em] \Rightarrow 12k(k + 1) - 8(k + 1) = 0 \\[1em] \Rightarrow (k + 1) - (12k - 8) = 0 \\[1em] \Rightarrow (k + 1)(12k - 8) = 0 \\[1em] \Rightarrow k + 1 = 0 \text{ or } 12k - 8 = 0 \\[1em] \Rightarrow k = -1 \text{ or } k = \dfrac{8}{12} \\[1em] \Rightarrow k = -1 \text{ or } k = \dfrac{2}{3}$

Hence, the value of k is -1, $\dfrac{2}{3}$.

(ii) The given equation is (k - 4)x² + 2(k - 4)x + 4 = 0.

Comparing with ax² + bx + c = we obtain,
a = k - 4 , b = 2(k - 4) , c = 4

$\therefore \text{Discriminant} = b^2 - 4ac \\[1em] = (2k - 8)^2 - 4 \times k - 4 \times 4 \\[1em] = 4k^2 + 64 - 32k - 16(k - 4) \\[1em] = 4k^2 - 32k - 16k + 64 + 64 \\[1em] = 4k^2 - 48k + 128$

For equal roots, discriminant = 0

$\Rightarrow 4k^2 - 48k + 128 = 0 \\[1em] \Rightarrow 4(k^2 - 12k + 32) = 0 \\[1em] \Rightarrow k^2 - 12k + 32 = 0 \\[1em] \Rightarrow k^2 - 8k - 4k + 32 = 0 \\[1em] \Rightarrow k(k - 8) - 4(k - 8) = 0 \\[1em] \Rightarrow k - 4 = 0 \text{ or } k - 8 = 0 \\[1em] \Rightarrow k = 4 \text{ or } k = 8 \\[1em]$

k ≠ 4 , as that will make a = (k - 4) = 0 and thus roots will become = ∞ .

Hence, the value of k is 8.

Find the value(s) of m for which each of the following quadratic equation has real and equal roots :

(i) (3m + 1)x² + 2(m + 1)x + m = 0

(ii) x² + 2(m - 1)x + (m + 5) = 0

Answer

(i) The given equation is (3m + 1)x² + 2(m + 1)x + m = 0.

Comparing with ax² + bx + c = we obtain,
a = 3m + 1 , b = 2(m + 1) , c = m

$\therefore \text{Discriminant} = b^2 - 4ac \\[0.5em] = (2m + 2)^2 - 4 \times 3m + 1 \times m \\[0.5em] = 4m^2 + 4 + 8m - 4m(3m + 1) \\[0.5em] = 4m^2 + 4 + 8m - 12m^2 - 4m \\[0.5em] = 4m^2 - 12m^2 + 8m - 4m + 4 \\[0.5em] = -8m^2 + 4m + 4 \\[0.5em]$

For equal roots, discriminant = 0

$\Rightarrow -8m^2 + 4m + 4 = 0 \\[0.5em] \Rightarrow -4(2m^2 - m - 1) = 0 \\[0.5em] \Rightarrow 2m^2 - m - 1 = 0 \\[0.5em] \Rightarrow 2m^2 - 2m + m - 1 = 0 \\[0.5em] \Rightarrow 2m(m - 1) + 1(m - 1) = 0 \\[0.5em] \Rightarrow (2m + 1)(m - 1) = 0 \\[0.5em] \Rightarrow 2m + 1 = 0 \text{ or } m - 1 = 0 \\[0.5em] \Rightarrow m = -\dfrac{1}{2} \text{ or } m = 1$

Hence, the value of m is $-\dfrac{1}{2}$ and 1.

(ii) The given equation is x² + 2(m - 1)x + (m + 5) = 0.

Comparing with ax² + bx + c = we obtain,
a = 1 , b = 2(m - 1) , c = m + 5

$\therefore \text{Discriminant} = b^2 - 4ac \\[0.5em] = (2m - 2)^2 - 4 \times 1 \times m + 5 \\[0.5em] = 4m^2 + 4 - 8m - 4(m + 5) \\[0.5em] = 4m^2 + 4 - 8m - 4m - 20 \\[0.5em] = 4m^2 - 12m - 16 \\[0.5em] = 4m^2 - 12m - 16 \\[0.5em]$

For equal roots, discriminant = 0

$\Rightarrow 4m^2 - 12m - 16 = 0 \\[0.5em] \Rightarrow 4(m^2 - 3m - 4) = 0 \\[0.5em] \Rightarrow m^2 - 3m - 4 = 0 \\[0.5em] \Rightarrow m^2 - 4m + m - 4 = 0 \\[0.5em] \Rightarrow m(m - 4) + 1(m - 4) = 0 \\[0.5em] \Rightarrow (m - 4)(m + 1) = 0 \\[0.5em] \Rightarrow m - 4 = 0 \text{ or } m + 1 = 0 \\[0.5em] \Rightarrow m = 4 \text{ or } m = -1$

Hence, the value of m is 4, -1.

Find the values of k for which each of the following quadratic equation has equal roots :

(i) 9x² + kx + 1 = 0

(ii) x² - 2kx + 7k - 12 = 0

Also, find the roots for those values of k in each case.

Answer

(i) The given equation is 9x² + kx + 1 = 0.

Comparing with ax² + bx + c = we obtain,
a = 9 , b = k , c = 1

$\therefore \text{Discriminant} = b^2 - 4ac \\[0.5em] = (k)^2 - 4 \times 9 \times 1 \\[0.5em] = k^2 - 36 \\[0.5em]$

For equal roots, discriminant = 0

$\Rightarrow k^2 - 36 = 0 \\[0.5em] \Rightarrow k^2 - (6)^2 = 0 \\[0.5em] \Rightarrow (k + 6)(k - 6) = 0 \\[0.5em] \Rightarrow k + 6 = 0 \text{ or } k - 6 = 0 \\[0.5em] \Rightarrow k = -6 \text{ or } k = 6 \\[0.5em]$

When k = 6 the equation becomes 9x² + 6x + 1 = 0 so the roots are :

The roots of the equation are given by

$x = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2a} \\[1em] \Rightarrow x = \dfrac{-(6) ± \sqrt{(6)^2 - 4\times 9 \times 1}}{2 \times 9} \\[1em] \Rightarrow x = \dfrac{-6 ± \sqrt{36 - 36}}{18} \\[1em] \Rightarrow x = \dfrac{-6 ± \sqrt{0}}{18} \\[1em] \Rightarrow x = \dfrac{-6 +0}{18} \text{ or } \dfrac{-6 - 0}{18} \\[1em] \Rightarrow x = -\dfrac{1}{3} \text{ or } -\dfrac{1}{3} \\[1em]$

When k = -6 the equation becomes 9x² - 6x + 1 = 0 so the roots are :

The roots of the equation are given by

$x = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2a} \\[1em] \Rightarrow x = \dfrac{-(-6) ± \sqrt{(-6)^2 - 4\times 9 \times 1}}{2 \times 9} \\[1em] \Rightarrow x = \dfrac{6 ± \sqrt{36 - 36}}{18} \\[1em] \Rightarrow x = \dfrac{6 ± \sqrt{0}}{18} \\[1em] \Rightarrow x = \dfrac{6 +0}{18} \text{ or } \dfrac{6 - 0}{18} \\[1em] \Rightarrow x = \dfrac{1}{3} \text{ or } \dfrac{1}{3} \\[1em]$

Hence, the values of k are 6, -6 ; when k = 6, roots are $-\dfrac{1}{3}, -\dfrac{1}{3}$ and when k = -6, roots are $\dfrac{1}{3}, \dfrac{1}{3}$.

(ii) The given equation is x² - 2kx + 7k - 12 = 0.

Comparing with ax² + bx + c = we obtain,
a = 1 , b = -2k , c = 7k - 12

$\therefore \text{Discriminant} = b^2 - 4ac \\[0.5em] = (-2k)^2 - 4 \times 1 \times (7k - 12) \\[0.5em] = 4k^2 - 4(7k - 12) \\[0.5em] = 4k^2 - 28k + 48$

For equal roots, discriminant = 0

$\Rightarrow 4k^2 - 28k + 48 = 0 \\[0.5em] \Rightarrow 4(k^2 - 7k + 12) = 0 \\[0.5em] \Rightarrow k^2 - 7k + 12 = 0 \\[0.5em] \Rightarrow k^2 - 4k - 3k + 12 = 0 \\[0.5em] \Rightarrow k(k - 4) - 3(k - 4) = 0 \\[0.5em] \Rightarrow (k - 3)(k - 4) = 0 \\[0.5em] \Rightarrow k - 3 = 0 \text{ or } k - 4 = 0 \\[0.5em] \Rightarrow k = 3 \text{ or } k = 4 \\[0.5em]$

When k = 3 the equation becomes x² - 6x + 9 = 0 so the roots are :

The roots of the equation are given by

$x = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2a} \\[1em] \Rightarrow x = \dfrac{-(-6) ± \sqrt{(-6)^2 - 4\times 1 \times 9}}{2 \times 1} \\[1em] \Rightarrow x = \dfrac{6 ± \sqrt{36 - 36}}{2} \\[1em] \Rightarrow x = \dfrac{6 ± \sqrt{0}}{12} \\[1em] \Rightarrow x = \dfrac{6 +0}{2} \text{ or } \dfrac{6 - 0}{2} \\[1em] \Rightarrow x = 3 \text{ or } 3 \\[1em]$

When k = 4 the equation becomes x² - 8x + 16 = 0 so the roots are :

The roots of the equation are given by

$x = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2a} \\[1em] \Rightarrow x = \dfrac{-(-8) ± \sqrt{(-8)^2 - 4\times 1 \times 16}}{2 \times 1} \\[1em] \Rightarrow x = \dfrac{8 ± \sqrt{64 - 64}}{2} \\[1em] \Rightarrow x = \dfrac{8 ± \sqrt{0}}{2} \\[1em] \Rightarrow x = \dfrac{8 + 0}{2} \text{ or } \dfrac{8 - 0}{2} \\[1em] \Rightarrow x = 4 \text{ or } 4 \\[1em]$

Hence the values of k are 3, 4 ; when k = 3, roots are 3, 3 and when k = 4, roots are 4, 4.

Find the value(s) of p for which the quadratic equation (2p + 1)x² - (7p + 2)x + (7p - 3) = 0 has equal roots. Also find these roots.

Answer

The given equation is (2p + 1)x² - (7P + 2)x + (7p - 3) = 0.

Comparing with ax² + bx + c = we obtain,
a = 2p + 1 , b = -(7p + 2) , c = 7p - 3

$\therefore \text{Discriminant} = b^2 - 4ac \\[1em] = (-(7p + 2))^2 - 4 \times (2p + 1) \times (7p - 3) \\[1em] = 49p^2 + 4 + 28p - 4(2p + 1)(7p - 3) \\[1em] = 49p^2 + 4 + 28p - 4(14p^2 - 6p + 7p - 3) \\[1em] = 49p^2 - 56p^2 + 28p - 4p + 4 + 12 \\[1em] = -7p^2 + 24p + 16$

For equal roots, discriminant = 0

$\Rightarrow -7p^2 + 24p + 16 \\[1em] \Rightarrow 7p^2 - 24p - 16 = 0 \text{ (Multiplying the equation by -1) } \\[1em] \Rightarrow 7p^2 - 28p + 4p - 16 = 0 \\[1em] \Rightarrow 7p(p - 4) + 4(p - 4) = 0 \\[1em] \Rightarrow (7p + 4)(p - 4) = 0 \\[1em] \Rightarrow 7p + 4 = 0 \text{ or } p - 4 = 0 \\[1em] \Rightarrow 7p = -4 \text{ or } p = 4 \\[1em] \Rightarrow p = -\dfrac{4}{7} \text{ or } p = 4 \\[1em]$

When p = $-\dfrac{4}{7}$ the equation becomes $(2 \times -\dfrac{4}{7} + 1)x^2 - (7 \times -\dfrac{4}{7} + 2)x + (7 \times -\dfrac{4}{7} - 3) = 0$ so the roots are :

$\Rightarrow -\dfrac{1}{7}x^2 + 2x - 7 = 0 \\[1em] \Rightarrow x^2 - 14x + 49 = 0 \text{ (On multiplying complete equation by -7) } \\[1em] \Rightarrow x^2 - 14x + 49 = 0 \\[1em] \Rightarrow x^2 - 7x - 7x + 49 = 0 \\[1em] \Rightarrow x(x - 7) - 7(x - 7) = 0 \\[1em] \Rightarrow (x - 7)(x - 7) = 0 \\[1em] \Rightarrow x - 7 = 0 \text{ or } x - 7 = 0 \\[1em] \Rightarrow x = 7 \text{ or } x = 7 \\[1em]$

When p = 4 the equation becomes $(2 \times 4 + 1)x^2 - (7 \times 4 + 2)x + (7 \times 4 - 3) = 0$ so the roots are :

$\Rightarrow 9x^2 - 30x + 25 = 0 \\[1em] \Rightarrow 9x^2 - 15x - 15x + 25 = 0 \\[1em] \Rightarrow 3x(3x - 5) - 5(3x - 5) = 0 \\[1em] \Rightarrow (3x - 5)(3x - 5) = 0 \\[1em] \Rightarrow 3x - 5 = 0 \text{ or } 3x - 5 = 0 \\[1em] \Rightarrow x = \dfrac{5}{3} \text{ or } x = \dfrac{5}{3} \\[1em]$

(Ans.) 4, $-\dfrac{4}{7}$ ; when p = $-\dfrac{4}{7}$, roots are 7, 7 and when p = 4, roots are $\dfrac{5}{3} , \dfrac{5}{3}.$

Find the value(s) of p for which the quadratic equation 2x² + 3x + p = 0 has real roots.

Answer

For real roots, Discriminant ≥ 0

or , b² - 4ac ≥ 0

The above equation is 2x² + 3x + p = 0
Comparing with ax² + bx + c = we obtain,
a = 2 , b = 3 , c = p

Putting values in b² - 4ac ≥ 0 we get,

$= 3^2 - 4 \times 2 \times p \ge 0 \\[0.5em] = 9 - 8p \ge 0 \\[0.5em] \Rightarrow 9 \ge 8p \\[0.5em] \Rightarrow 8p \le 9 \\[0.5em] \Rightarrow p \le \dfrac{9}{8} \\[0.5em]$

Hence the value of p is ≤ $\dfrac{9}{8}$.

Find the least positive value of k for which the equation x² + kx + 4 = 0 has real roots.

Answer

For real roots, Discriminant ≥ 0

or , b² - 4ac ≥ 0

The above equation is x² + kx + 4 = 0
Comparing with ax² + bx + c = we obtain,
a = 1 , b = k , c = 4

Putting values in b² - 4ac ≥ 0 we get,

$= k^2 - 4 \times 1 \times 4 \ge 0 \\[0.5em] = k^2 - 16 \ge 0 \\[0.5em] \Rightarrow k^2 - 4^2 \ge 0 \\[0.5em] \Rightarrow (k - 4)(k + 4) \ge 0 \\[0.5em] \Rightarrow k \ge 4 \\[0.5em]$

Hence least positive value for which equation has real roots is 4 .

Find the values of p for which the equation 3x² - px + 5 = 0 has real roots.

Answer

For real roots,

Discriminant ≥ 0

or, b² - 4ac ≥ 0

The above equation is 3x² - px + 5 = 0
Comparing with ax² + bx + c = we obtain,
a = 3 , b = -p , c = 5

Putting values in b² - 4ac ≥ 0 we get,

$= (-p)^2 - 4 \times 3 \times 5 \ge 0 \\[1em] = p^2 - 60 \ge 0 \\[1em] \Rightarrow p^2 - (\sqrt{60})^2 \ge 0 \\[1em] \Rightarrow (p + \sqrt{60})(p -\sqrt{60}) \ge 0 \\[1em] \Rightarrow (p + 2\sqrt{15})(p - 2\sqrt{15}) \ge 0 \\[1em] \text{(Using) } (a + b)(a - b) \ge 0 = a \ge b \text{ or } a \le -b \text{ we get, } \\[1em] \Rightarrow p \ge 2\sqrt{15} \text{ or } p \le -2\sqrt{15} \\[1em]$

Hence the values of p are $p \le -2\sqrt{15} \text{ or } p\ge 2\sqrt{15}$ .