Question

Find all real numbers k for which the equation (k-5)x^2-kx+5=0 has exactly one real solution.

Answer 1

The equation  $k^2-20k+100=0$ will have exactly one real solution when k=10.

Here, you will find k from the value of the discriminant for the given quadratic equation.

The standard form for a quadratic equation is: $ax^{2}+bx+x=0$. Where: a, b, and c are your coefficients.

Then, the given equation $(k-5)x^2-kx+5=0$  is a quadratic equation.  Where:

                                                  $a=k-5\\ \\ b=-k\\ \\ c=5$

The number of solutions for a quadratic equation is determined by the discriminant (D) that can be calculated from the expression: $b^2-4a*c$. When:

• D > 0 - the equation has two solutions;
• D = 0 -  the equation has only one solution;
• D < 0 - there are no solutions for the equation.

• STEP 1 - Calculate the discriminant for the given equation.

                                       $D=b^2-4a*c\\ \\ D=(-k)^2-4*(k-5)*5\\ \\ D=k^2-20*(k-5)\\ \\ D=k^2-20k+100$

• STEP 2 - Find k.

You should find the value for k for which the given equation has only one real solution. Therefore, D=0. For this, solve the quadratic formula for the equation  $D=k^2-20k+100=0$ .

                                    $k_{1,2}=\frac{\(-b\pm \sqrt{(b)^2-4\cdot \:a\cdot \:c}}{2\cdot \:a}\\\ \\ k_{1,2}=\frac{\(20\pm \sqrt{(-20)^2-4\cdot \:1\cdot \:100}}{2\cdot \:1}\\ \\ k_{1,\:2}=\frac{20\pm \sqrt{0}}{2\cdot \:1}\\ \\ k=\frac{20}{2}=10$

Hence, k=10 for the equation  $k^2-20k+100=0$ has exactly one real solution.

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