Question

Solve by using quadratic formula, adx^2 - x (ab - cd) = bc.​

Answer 1

$\mathsf{a d x ^ { 2 } - x ( a b - c d ) = b c}$

• Use the distributive property to multiply x by ab-cd.

$\mathsf{adx^{2}-\left(xab-xcd\right)=bc }$

• To find the opposite of xab-xcd, find the opposite of each term.

$\mathsf{adx^{2}-abx=-cdx+bc }$

• Reorder the terms.

$\mathsf{adx^{2}-abx=-cdx+bc }$

• $\mathsf{a=-\frac{c}{x} }$

Answer 2

${\huge{\underline{\boxed{\red{\mathcal{Answer}}}}}}$

Given:

• $adx^2 - x(ab - cd) = bc$

To Find:

• Value of x

Solution:

$\implies adx^2 - x(ab - cd) = bc \\\\\implies adx^2 - (ab - cd)x - bc = 0 \\\\\implies adx^2 + (cd - ab)x - bc = 0 \\\\\implies x = \dfrac{-A \pm \sqrt{B^2 - 4AC\:}}{2A} \;\;\;\;Where, A=ad,\:B=(cd-ab),\:C=-bc \\\\\implies x = \dfrac{-(cd-ab) \pm \sqrt{(cd-ab)^2 - 4(ad)(-bc)}\:}{2(ad)} \\\\\implies x = \dfrac{(ab-cd) \pm \sqrt{(c^2d^2 + a^2d^2 - 2abcd) + 4abcd\:}}{2ad} \\\\\implies x = \dfrac{(ab-cd) \pm \sqrt{c^2d^2 + a^2d^2 - 2abcd + 4abcd\:}}{2ad} \\\\\implies x = \dfrac{(ab-cd) \pm \sqrt{c^2d^2 + a^2d^2 + 2abcd\:}}{2ad} \\\\\implies x = \dfrac{(ab-cd) \pm \sqrt{(cd + ab)^2\:}}{2ad}\\\\\implies x = \dfrac{(ab-cd) \pm (cd + ab)}{2ad} \\\\\implies x = \dfrac{(ab - cd) + (cd + ab)}{2ad} \:\:\:\ OR \:\:\:x = \dfrac{(ab - cd) - (cd + ab)}{2ad} \\\\\implies x = \dfrac{ab - cd\!\!\!/ + cd\!\!\!/ + ab}{2ad} \:\:\:\ OR \:\:\:x = \dfrac{ab\!\!\!/ - cd - cd - ab\!\!\!/ }{2ad} \\\\\implies x = \dfrac{2\!\!/\:\:*a\!\!/\:\:*b}{2\!\!/\:\:*a\!\!/\:\:*d} \:\:\:\ OR \:\:\:x = \dfrac{-2\!\!/\:\:*c*d\!\!/\:\:}{2\!\!/\:\:*a*d\!\!/\:\:} \\\\\bold{\implies x = \dfrac{b}{d} \:\:\:\:\: OR\:\:\: x = \dfrac{-c}{a} }$

Formula used:

• $x = \dfrac{-A \pm \sqrt{B^2 - 4AC\:}}{2A}$
• $(p-q)^2 = p^2 + q^2 - 2pq$
• $p^2 + q^2 + 2pq = (p+q)^2$

Hope it helps!!