Example Question - solving square root equation
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Solving Square Root Equation with Quadratic Formula
L'image est floue, mais je peux distinguer suffisamment pour répondre à la question. La question semble demander de résoudre une équation avec des racines carrées. L'équation est : \( 4x + 8(\sqrt{x}) - 5 = 0 \) Pour résoudre cette équation, on peut poser \( t = \sqrt{x} \), alors \( t^2 = x \). L'équation devient : \( 4t^2 + 8t - 5 = 0 \) C'est maintenant une équation du second degré. On peut utiliser la formule quadratique pour trouver les valeurs de \( t \), qui est : \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) Où \( a = 4 \), \( b = 8 \), et \( c = -5 \). Donc, \( t = \frac{-8 \pm \sqrt{8^2 - 4*4*(-5)}}{2*4} \) \( t = \frac{-8 \pm \sqrt{64 + 80}}{8} \) \( t = \frac{-8 \pm \sqrt{144}}{8} \) \( t = \frac{-8 \pm 12}{8} \) On obtient alors deux solutions possibles pour \( t \) : \( t_1 = \frac{4}{8} = \frac{1}{2} \) \( t_2 = \frac{-20}{8} = -\frac{5}{2} \) Cependant \( t \) étant égal à \(\sqrt{x}\), il ne peut pas être négatif, donc nous rejetons la solution \( t_2 = -\frac{5}{2} \). Nous sommes donc laissés avec \( t = \frac{1}{2} \), ce qui implique que \( \sqrt{x} = \frac{1}{2} \). En élevant les deux côtés au carré, nous obtenons \( x = \left(\frac{1}{2}\right)^2 \), soit \( x = \frac{1}{4} \). La valeur approchée de \( x \) sera donc: \( x ≈ \frac{1}{4} \)
Solving Square Root Equation
The equation provided in the image is \( \sqrt{x} - 2 - 1 = 0 \) To solve for x, follow these steps: 1. Combine the constants on the left side: \( \sqrt{x} - 3 = 0 \) 2. Isolate the square root: \( \sqrt{x} = 3 \) 3. Square both sides to eliminate the square root: \( (\sqrt{x})^2 = 3^2 \) \( x = 9 \) The solution to the equation is \( x = 9 \).
Solving a Square Root Equation
The mathematical expression you provided is difficult to read due to the low resolution of the image, but here is what it seems to be: √(x+7) - √(x-2) = √(x+2) - √(x-2) - (5/2) To solve for x, one would typically isolate the square roots on one side and then square both sides to eliminate the square root terms. However, there seems to be a mistake, because both sides of the equation contain the term √(x-2), which would cancel out. If this is correct, the equation after canceling the √(x-2) terms would simplify to: √(x+7) = √(x+2) - (5/2) Now let's proceed with this corrected equation. First, isolate the square root on one side: √(x+7) + (5/2) = √(x+2) Then square both sides to get rid of the square roots: (√(x+7) + 5/2)^2 = (√(x+2))^2 Expand the left-hand side: (x+7) + 2*(5/2)*√(x+7) + (5/2)^2 = x + 2 Simplify further by squaring (5/2) and combining like terms: x + 7 + 5√(x+7) + 25/4 = x + 2 Now, isolate the square root term: 5√(x+7) = x + 2 - (x + 7) - 25/4 5√(x+7) = -5 - 25/4 Since we cannot have a square root equal to a negative number (when considering real numbers), it seems there is no solution to the equation in the real number system. Please check the original equation to ensure it has been transcribed correctly. If there is a different equation, please provide a clearer image or the correct terms.
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