Question

for which value of k will the roots of the equation $2x^2 - 4x + k = 0$ be real and equal?
(1) 1 (2) 2 (3) -1 (4) 0
your answer

what are the solutions to the equation
$3x^2 + 3 = 7x$
(1) $x = \frac{7 + \sqrt{85}}{6}$ or $x = \frac{7 - \sqrt{85}}{6}$ (3) $x = \frac{7 + \sqrt{13}}{6}$ or $x = \frac{7 - \sqrt{13}}{6}$
(2) $x = \frac{-7 + \sqrt{85}}{6}$ or $x = \frac{-7 - \sqrt{85}}{6}$ (4) $x = \frac{-7 + \sqrt{13}}{6}$ or $x = \frac{-7 - \sqrt{13}}{6}$

Explanation:

First Question:

Step1: Recall discriminant condition

For quadratic $ax^2+bx+c=0$, real equal roots need $\Delta = b^2-4ac=0$

Step2: Identify coefficients

$a=2$, $b=-4$, $c=k$

Step3: Substitute into discriminant

$\Delta = (-4)^2 - 4(2)(k) = 0$

Step4: Solve for k

$16 - 8k = 0 \implies 8k=16 \implies k=2$

Step1: Rearrange to standard form

$3x^2 -7x +3 = 0$

Step2: Identify coefficients

$a=3$, $b=-7$, $c=3$

Step3: Apply quadratic formula

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Step4: Compute discriminant

$\Delta = (-7)^2 -4(3)(3)=49-36=13$

Step5: Substitute into formula

$x=\frac{7\pm\sqrt{13}}{6}$

Answer:

(2) 2

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Second Question: