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Question: How do you solve \(2{x^2} - 13x + 20 = 0\)?...

How do you solve 2x213x+20=02{x^2} - 13x + 20 = 0?

Explanation

Solution

This question is from the topic of quadratic equations. In this question we need to find the roots of a given quadratic equation 2x213x+20=02{x^2} - 13x + 20 = 0. We can find the root of the quadratic equation by completing the square method, factor method and quadratic formula. We will use a quadratic formula to solve the given quadratic equation, because it is easiest.

Complete step by step solution:
Let us try to solve this question in which we need to find the roots of a
given quadratic equation 2x213x+20=02{x^2} - 13x + 20 = 0. We are using a quadratic formula to find the roots given the equation. Quadratic formula is given by b±b24ac2a\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}for any general quadratic equation ax2+bx+c=0a{x^2} + bx + c = 0 where b24ac{b^2} - 4ac is called the discriminant of quadratic equation, it tells the nature of roots of quadratic equation. Here are conditions:

  1. Two distinct real roots, if b24ac>0{b^2} - 4ac > 0
  2. Two equal real roots, if b24ac=0{b^2} - 4ac = 0
  3. No real roots, if b24ac<0{b^2} - 4ac < 0
    In the given quadratic equation we have,
    a=2 b=13 c=20  a = 2 \\\ b = - 13 \\\ c = 20 \\\
    Discriminant of the quadratic equation is
    b24ac=(13)24220 =169160 =9>0   {b^2} - 4ac = {(13)^2} - 4 \cdot 2 \cdot 20 \\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 169 - 160 \\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 9 > 0 \\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\\
    Hence the given quadratic equation has two distinct real roots, because discriminant is greater than 00.
    Now putting values of a,ba,b and cc in quadratic formula we get,
x=b±b24ac2a=(13)±(13)2424022 =13±1691604 =13±94 x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} = \dfrac{{ - ( - 13) \pm \sqrt {{{( - 13)}^2} - 4 \cdot 2 \cdot 40} }}{{2 \cdot 2}} \\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{{13 \pm \sqrt {169 - 160} }}{4} \\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{{13 \pm \sqrt 9 }}{4} \\\

As we know that square root of 99is equal to±3 \pm 3. So we get,
x=13±34x = \dfrac{{13 \pm 3}}{4}
x=13+34x = \dfrac{{13 + 3}}{4} And x=1334x = \dfrac{{13 - 3}}{4}
x=164=4x = \dfrac{{16}}{4} = 4 And x=104=52x = \dfrac{{10}}{4} = \dfrac{5}{2}
Hence the root of the quadratic equation 2x213x+20=02{x^2} - 13x + 20 = 0 are x=4x = 4andx=52x = \dfrac{5}{2}.

Note: To solve questions in which you are asked to find the roots of quadratic equations by quadratic formula you must need to know the formula. We can also solve this by using other methods of finding roots of a quadratic equation such as completing square method and factor method.