Question

If the p and q are the roots of the equation 2x ^ 2 - 4x - 3 = 0 the value of p ^ 2 + q ^ 2 is (a) 5 (b)7 (c)3; (d) - 4​

Answer 1

EXPLANATION.

p and q are the roots of the equation.

⇒ 2x² - 4x - 3 = 0.

As we know that,

Sum of the zeroes of the quadratic expression.

⇒ p + q = - b/a.

⇒ p + q = -(-4)/2 = 2.

Products of the zeroes of the quadratic expression.

⇒ pq = c/a.

⇒ pq = (-3)/2 = - 3/2.

To find :

⇒ p² + q².

As we know that,

We can write equation as,

⇒ p² + q² = [p + q]² - 2pq.

Put the values in the equation, we get.

⇒ p² + q² = (2)² - 2(-3/2).

⇒ p² + q² = 4 + 3.

⇒ p² + q² = 7.

Option [B] is correct answer.

                                                                                                                         

MORE INFORMATION.

Nature of the roots of the quadratic expression.

(1) = Real and unequal, if b² - 4ac > 0.

(2) = Rational and different, if b² - 4ac is a perfect square.

(3) = Real and equal, if b² - 4ac = 0.

(4) = If D < 0 Roots are imaginary and unequal Or complex conjugate.

Answer 2

$\bf \underline{Question : }$

If the p and q are roots of the equation

$2 {x}^{2}$ - 4x - 3 = 0 . The value of $\sf{p}^{2} + {q}^{2}$is :-

(a) 5

(b) 7 ✓

(c) 3

(d) - 4

$\bf \underline{Answer : }$

$\sf b) 7$

$\bf \underline{Explanation : }$

$\sf \underline{Given \: Equation : }$

$\tt2 {x}^{2} - 4x - 3 = 0$

$\sf We \: know \: that \begin{cases} \sf p +q = \frac{ - b}{ \: \: a} \\ \\ \sf pq = \frac{c}{a} \end{cases}$

$\sf where \begin{cases} \sf a = \: \: \: 2 \\ \sf b = - 4 \\ \sf c = - 3 \end{cases}$

$\sf \underline{Sum \: of \: roots }$

$\sf p + q = \frac{ \cancel- (\cancel-4 )}{2} = \frac{4}{2} = 2$

$\sf \underline{Product \: of \: roots }$

$\sf pq = \frac{ - 3}{ \: \: 2}$

$\sf {p}^{2} + {q}^{2} = {(p + q)}^{2} - 2pq$

$\sf Putting \: values \: in \: above \: equation$

$\sf \implies {p}^{2} + {q}^{2} = {(2)}^{2} - 2( \frac{ - 3}{ \: \: 2} )$

$\sf \implies {p}^{2} + {q}^{2} = 4 + 3 \\ \sf \implies {p}^{2} + {q}^{2} =7 \: \: \: \: \: \: \:$