How can we make it easy to solve quadratics without the Quadratic Formula?For example,5x2−7x−2=0Then, we only need to find is when5x2−7x=2x(5x−7)=2Since that when we get 2−2=0.Is there any way we can find them easily without solving quadratics using formula?

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Jadyn Gentry · Accepted Answer

Unfortunately, your strategy does not help that much because it is very difficult to solvex(5x−7)=2without simply returning to the original equation. This is because there is a 2 on the RHS. The reason we want to be able to write a quadratic in the form(x−p)(x−q)=0is so that we can exploit the zero-product property: if&amp;nbsp;a×b=0, then&amp;nbsp;a=0&amp;nbsp;or&amp;nbsp;b=0. Once we have written an equation in the form&amp;nbsp;(x−p)(x−q)=0, we can still instantly deduce that&amp;nbsp;x=p&amp;nbsp;or&amp;nbsp;x=q. This is something we can&#039;t do when we have something likex(5x−7)=2,.Instead, I would suggest dividing through by 5, which yieldsx2−75x−25=0,.Then, add 2/5 to both sides:x2−75x=25&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;(*)&amp;nbsp;The LHS is almost a perfect square. Note that(x−710)2=x2−75x+49100,.If we subtract 49/100 from this equation, we get(x−710)2−49100=x2−75x,.Hence, the equation (*) can be rewritten as(x−710)2−49100=25,.Add 49/100 to both sides:(x−710)2=25+49100=89100,.Take the square root of both sides:x−710=±89100,.Add 7/10 to both sides:x=710±89100,,and we are done. This method of solving quadratics is known as completing the square. Unlike factorisation, it can be used to solve any quadratic. In fact, the quadratic formula comes from completing the square on the general quadratic equationax2+bx+c=0,.

Marquis Ibarra · Accepted Answer

All ways of solving quadratic equations are essentially the same, but here&#039;s another tack:You can &quot;depress&quot; the quadratic ax2+bx+c using the substitution x=y−b2a. This substitution makes the linear term disappear. Then it&#039;s easy to solve for y and then plug that answer into the substitution and recover x.For your example: 5x2−7x−2=0, let x=y−−72⋅5=y+710. The equation becomes 5(y+710)2−7(y+710)−2=0which simplifies to5y2−8920=0.Solve for y (which is easy because there&#039;s no linear term):y2=89100y=±8910.Which meansx=±8910+710.So all the steps are easy and obvious as long as you remember the substitution x=y−b2a.

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