Question

Using quadratic formula find the roots of (2x+3)(3x-7)=0​

Answer 1

EXPLANATION.

Quadratic formula = (2x + 3)(3x - 7).

As we know that,

Expand the equation.

⇒ 2x[3x - 7] + 3[3x - 7].

⇒ 6x² - 14x + 9x - 21.

⇒ 6x² - 5x - 21.

As we know that,

Sum of the zeroes of the quadratic equation.

⇒ α + β = -b/a.

⇒ α + β = -(-5)/6 = 5/6.

Products of the zeroes of the quadratic equation.

⇒ αβ = c/a.

⇒ αβ = -21/6 = -7/2.

Equation of quadratic polynomial.

⇒ x² - (α + β)x + αβ.

Put the values in equation, we get.

⇒ x² - (5/6)x + (-7/2) = 0.

⇒ 6x² - 5x - 21 = 0.

HENCE PROVED.

                                                                                                                                           

MORE INFORMATION.

Nature of the factors of the quadratic expression.

(1) = Real and different, 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

Answer:

$\tt{ \dfrac{-3}{2}\:and\:\dfrac{7}{3}}$

Step-by-step explanation:

Question:

$\tt{Find\:the \:roots \:of \:(2x+3)(3x-7)=0}$

Answer:

$\tt{(2x+3)(3x-7)=0}$

$\implies\tt{2x+3=0\:and\:3x-7=0}$

$\implies\tt{2x= -3\:and\:3x= 7}$

$\implies\tt{x= \dfrac{-3}{2}\:and\:x= \dfrac{7}{3}}$

Hence the roots are: $\tt{ \dfrac{-3}{2}\:and\:\dfrac{7}{3}}$