Advertisements
Advertisements
प्रश्न
Find the value of ‘K’ for which x = 3 is a solution of the quadratic equation, (K + 2)x2 – Kx + 6 = 0. Also, find the other root of the equation.
उत्तर
(K + 2)x2 – Kx + 6 = 0 …(1)
Substitute x = 3 in equation (1)
(–4 + 2)x2 –(–4)x + 6 = 0
⇒ –2x2 + 4x + 6 = 0
⇒ x2 – 2x – 3 = 0 ...(Dividing by 2)
⇒ x2 – 3x + x – 3 = 0
⇒ x(x – 3) + 1(x – 3) = 0
(x + 1)(x – 3) = 0
So, the roots are x = –1 and x = 3
Thus, the other root of the equation is x = –1.
APPEARS IN
संबंधित प्रश्न
If (x + 2) and (x + 3) are factors of x3 + ax + b, find the values of 'a' and `b'.
Find the values of constants a and b when x – 2 and x + 3 both are the factors of expression x3 + ax2 + bx – 12.
Using the Factor Theorem, show that (3x + 2) is a factor of 3x3 + 2x2 – 3x – 2. Hence, factorise the expression 3x3 + 2x2 – 3x – 2 completely.
(3x + 5) is a factor of the polynomial (a – 1)x3 + (a + 1)x2 – (2a + 1)x – 15. Find the value of ‘a’, factorise the given polynomial completely.
Using the factor Theorem, show that:
2x + 7 is a factor 2x3 + 5x2 − 11x – 14. Hence, factorise the given expression completely.
If x - 2 and `x - 1/2` both are the factors of the polynomial nx2 − 5x + m, then show that m = n = 2
Prove by factor theorem that
(x - 3) is a factor of 5x2 - 21 x +18
Show that (x – 3) is a factor of x3 – 7x2 + 15x – 9. Hence factorise x3 – 7x2 + 15 x – 9
Using factor theorem, show that (x – 5) is a factor of the polynomial
2x3 – 5x2 – 28x + 15
x – 1 is a factor of 8x2 – 7x + m; the value of m is ______.
