Question

Find the values of m for which ${{x}^{2}}-m(2x-8)-15=0$ has
( a ) equal roots
( b ) both positive roots

Answer 1

To solve this question what we will do is we will put discriminant of given quadratic equation ${{x}^{2}}-m(2x-8)-15=0$ equals to 0 as we have two equal roots when ${{b}^{2}}-4ac=0$, then we have two equal roots. And then we will check that for which value of m we have two positive roots.

Now before we solve this question le us see what are the various conditions for the roots of the quadratic equation.
Let, we have a general quadratic equation of form $a{{x}^{2}}+bx+c=0$ , then the nature of roots are determined by value of discriminant which is ${{b}^{2}}-4ac$ .
So, if ${{b}^{2}}-4ac>0$, then we have two real roots.
If ${{b}^{2}}-4ac=0$, then we have two equal roots.
If ${{b}^{2}}-4ac<0$, we have two imaginary roots.
So, now move to the question. it is given in the question that we have to find the value of such that quadratic equation ${{x}^{2}}-m(2x-8)-15=0$has two equal roots, which means the discriminant value of the given quadratic equation should be equal to zero.
On comparing ${{x}^{2}}-m(2x-8)-15=0$with general quadratic equation$a{{x}^{2}}+bx+c=0$, we get
a = 1 , b = - 2m and c = 8m - 15
so, discriminant of ${{x}^{2}}-m(2x-8)-15=0$ will be equal to ${{\left( -2m \right)}^{2}}-4\left( 1 \right)(8m-15)$
for equal roots, ${{\left( -2m \right)}^{2}}-4\left( 1 \right)(8m-15)=0$
$4{{m}^{2}}-4(8m-15)=0$
On simplifying we get
$4{{m}^{2}}-32m+60=0$
${{m}^{2}}-8m+15=0$
Using quadratic formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ to get, value of m, we get
$m=\dfrac{-(-8)\pm \sqrt{{{(-8)}^{2}}-4(1)(15)}}{2(1)}$
$m=\dfrac{8\pm \sqrt{64-60}}{2}$
$m=\dfrac{8\pm 2}{2}$
On Solving, we get
$m=5,3$
So, at $m=5,3$we will get equal roots for quadratic equation ${{x}^{2}}-m(2x-8)-15=0$
Now, for both positive roots, condition can be defined as $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}>0$
Now, on re – writing ${{x}^{2}}-m(2x-8)-15=0$, we get
${{x}^{2}}-2mx+8m-15=0$
Or, ${{x}^{2}}-2mx+(8m-15)=0$.
On comparing, ${{x}^{2}}-2mx+(8m-15)=0$with $a{{x}^{2}}+bx+c=0$, we get
a = 1, b = - 2m, c = 8m – 15.
So, now putting values of a, b and c in $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}>0$, we get
$\dfrac{-(-2m)\pm \sqrt{{{(-2)}^{2}}-4(1)(8m-15)}}{2(1)}>0$
$\dfrac{2m\pm \sqrt{4-4(8m-15)}}{2}>0$
Taking 4 out of square root, we get
$\dfrac{2m\pm 2\sqrt{1-(8m-15)}}{2}>0$
$m\pm \sqrt{16-8m}>0$
So, we get$m+\sqrt{16-8m}>0$and $m-\sqrt{16-8m}>0$
Solving $m+\sqrt{16-8m}>0$
$m>-\sqrt{16-8m}$
Squaring both sides, we get
${{m}^{2}}>16-8m$
Or, ${{m}^{2}}+8m-16>0$
Using, formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ for finding value of m, we get
$m=\dfrac{-8\pm \sqrt{{{8}^{2}}-4(1)(-16)}}{2(1)}$
On solving, we get
$m=\dfrac{-8\pm \sqrt{128}}{2}$
$m=-4\pm 4\sqrt{2}$
Solving $m-\sqrt{16-8m}>0$
$m>\sqrt{16-8m}$
Squaring both sides, we get
${{m}^{2}}>16-8m$
Or, ${{m}^{2}}+8m-16>0$
Using, formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ for finding value of m, we get
$m=\dfrac{-8\pm \sqrt{{{8}^{2}}-4(1)(-16)}}{2(1)}$
On solving, we get
$m=\dfrac{-8\pm \sqrt{128}}{2}$
$m=-4\pm 4\sqrt{2}$
So, for $m=-4\pm 4\sqrt{2}$, we have two positive roots for ${{x}^{2}}-2mx+(8m-15)=0$.

Note: While solving the questions based on quadratic equation, one must know the quadratic formula for finding roots which is $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ for quadratic equation $a{{x}^{2}}+bx+c=0$. Also, conditions for nature of roots must also be remembered and also calculation mistakes should be avoided. For both positive roots, condition can be defined as $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}>0$.