Find all real numbers $z$ $z$ for which the equation $(z - 5) x^{2} - z x + 5 = 0$ $(z-5)x^2-zx+5=0$ only has one real solution. Help, Please?

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Answer 1 · 2018-01-25T03:08:43

$\Rightarrow z = 10 \pm 4 \sqrt{5}$ $=>z=10+-4sqrt(5)$

In order for a quadratic equation in the form $a x^{2} + b x + c = 0$ $ax^2+bx+c=0$ to have 1 solution, $b^{2} - 4 a c = 0$ $b^2-4ac=0$

In this case:

plug in the values to get:

${(- z)}^{2} - 4 (z - 5) (5) = 0$ $(-z)^2-4(z-5)(5)=0$

$\Rightarrow {(- z)}^{2} - 4 (5 z - 5) = 0$ $=>(-z)^2-4(5z-5)=0$

$\Rightarrow z^{2} - 20 z + 20 = 0$ $=>z^2-20z+20=0$

$\Rightarrow z = \frac{- (- 20) \pm \sqrt{{(- 20)}^{2} - 4 (1) (20)}}{2 \cdot 1}$ $=>z=(-(-20)+-sqrt((-20)^2-4(1)(20)))/(2*1)$

$\Rightarrow z = \frac{20 \pm \sqrt{400 - 80}}{2}$ $=>z=(20+-sqrt(400-80))/2$

$\Rightarrow z = \frac{20 \pm \sqrt{320}}{2}$ $=>z=(20+-sqrt(320))/2$

$\Rightarrow z = \frac{20 \pm 8 \sqrt{10}}{2}$ $=>z=(20+-8sqrt(10))/2$

$\Rightarrow z = 10 \pm 4 \sqrt{5}$ $=>z=10+-4sqrt(5)$

Find all real numbers zz for which the equation (z−5)x2−zx+5=0(z-5)x^2-zx+5=0 only has one real solution. Help, Please?